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A  TREATISE 

ON 

CHEMISTRY  AND  CHEMICAL 
ANALYSIS 

PREPARED  FOR  STUDENTS  OF 

THE  INTERNATIONAL  CORRESPONDENCE  SCHOOLS 

^  SCRANTON,  PA. 


Volume  I 


ARITHMETIC 

ELEMENTARY  ALGEBRA  AND  TRIGONOMETRIC 

FUNCTIONS 

PHYSICS 

THEORETICAL  CHEMISTRY 

WITH    PRACTICAL   QUESTIONS   AND    EXAMPLES 


First  Edition 


SCRANTON 

THE  COLLIERY  ENGINEER  CO. 
1900 

101 


i  ( 


Copyright^  1897, 1898, 1899,  by  THE  COLLIERY  ENGINEER  COMPANY. 


Arithmetic:   Copyright,  1892,  1893,  1894,  1895,  1896,  1898,  1899,   by   THE  COLLIERY 

ENGINEER  COMPANY. 
Elementary  Algebra  and  Trigonometric  Functions :  Copyright,  1893,  1894,  1895, 

1896,  1897,  1899,  by  THE  COLLIERY  ENGINEER  COMPANY. 
Physics :  Copyright,  1898,  by  THE  COLLIERY  ENGINEER  COMPANY. 
Theoretical  Chemistry :  Copyright,  1898,  by  THE  COLLIERY  ENGINEER  COMPANY. 


Press  of  EATON  &  MAINS 
NEW  YORK 


PREFACE. 


In  the  first  five  of  the  six  volumes  composing  this  set  are 
included  all  the  Instruction  and  Question  Papers  used  in  our 
Course  in  Chemistry,  including  Qualitative  and  Quantitative 
Analysis ;  they  give  in  a  clear  and  concise  form  not  only  the 
theory,  but  also  the  practical  applications  of  chemistry  and 
chemical  analysis.  These  volumes  constitute  a  valuable  work 
of  reference  in  connection  with  the  innumerable  problems 
so  frequently  met  with  in  analytical  chemistry  and  in  the 
manufacturing  industries. 

The  subject  of  Inorganic  Chemistry  is  very  thoroughly 
covered,  and  we  have  introduced  a  number  of  simple  experi- 
ments that  will  serve  to  illustrate  the  text  and  render  the 
student  familiar  with  the  manipulation  and  use  of  the 
apparatus  most  frequently  employed. 

That  portion  of  the  Course  dealing  with  Organic  Chem- 
istry has  been  so  arranged  as  to  render  evident  the  relations 
existing  between  the  various  classes  of  compounds;  for  this 
reason  the  usual  practice  of  treating  the  fatty  and  aromatic 
derivatives  separately  has  not  been  adopted,  the  compounds 
of  both  classes  being  considered  together  in  their  appropriate 
places. 

The  papers  treating  on  Qualitative  and  Quantitative 
Analysis  have  been  prepared  with  great  care;  the  student 
who  thoroughly  studies  the  text  and  performs  the  analyses 
required  will  be  able  to  make  all  the  analyses  usually  per- 
formed by  analytical  chemists. 

The  method  of  numbering  the  pages,  cuts,  articles,  etc.  is 
such  that  each  paper  and  part  is  complete  in  itself;  hence, 
in  order  to  make  the  indexes  intelligible,  it  was  necessary 

iii 


991174 


iv  PREFACE. 

to  give  each  paper  and  part  a  number.  This  number  is 
placed  at  the  top  of  each  page,  on  the  headline,  opposite  the 
page  number;  and  to  distinguish  it  from  the  page  number,  it 
is  preceded  by  the  printer's  section  mark  (§).  Consequently, 
a  reference  such  as  page  29,  §  18,  would  be  readily  found  as 
follows:  The  back  stamp  on  Vols.  I-V  shows  the  sections 
(i.  e.,  papers)  included  in  the  volume,  that  for  Vol.  V  reading 
§§  16-19 ;  hence,  look  in  Vol.  V  along  the  headlines  until  §  18 
is  found,  and  then  through  §  18  until  page  29  is  found. 

The  Question  Papers  are  given  the  same  section  numbers 
as  the  Instruction  Papers  to  which  they  belong,  and  are 
grouped  together  at  the  end  of  the  volumes  containing  the 
Instruction  Papers  to  which  they  refer. 

The  volumes  for  the  present  Course,  Chemistry,  including 
Qualitative  and  Quantitative  Analysis,  are  six  in  number: 

Vol.  I  (§§  1-5)  contains  the  Instruction  and  Question 
Papers  on  Arithmetic,  Elementary  Algebra  and  Trigonomet- 
ric Functions,  Physics,  and  Theoretical  Chemistry. 

Vol.  II  (§§  6-9)  contains  the  Instruction  and  Question 
Papers  on  Inorganic  Chemistry,  Parts  1-4. 

Vol.  Ill  (§§  10  and  11)  contains  the  Instruction  and  Ques- 
tion Papers  on  Qualitative  Analysis,  Parts  1  and  2. 

Vol.  IV  (§§  12-15)  contains  the  Instraction  and  Question 
Papers  on  Organic  Chemistry,  Parts  1-4. 

Vol.  V  (§§  16-19)  contains  the  Instruction  and  Question 
Papers  on  Quantitative  Analysis,  Parts  1-4. 

Vol.  VI  contains  the  answers  to  the  questions  and  solutions 
to  the  examples  in  the  Question  Papers.  Whenever  it  has 
been  deemed  inadvisable  to  answer  a  question,  a  reference 
to  the  proper  article  in  the  Instruction  Paper  has  been  given, 
the  reading  of  which  will  enable  the  student  to  answer  the 
question  himself.  In  a  number  of  cases  the  student  is  given 
a  solution  or  some  solid  substance  to  analyze.  While  these 
analyses  constitute  a  part  of  the  student's  work  and  form  a 
part  of  the  Question  Paper,  it  is  not  possible  to  give  the 
answers,  owing  to  the  fact  that  the  solutions  and  solids  are 
varied  from  time  to  time. 

THE  INTERNATIONAL  CORRESPONDENCE  SCHOOLS. 


CONTENTS. 


ARITHMETIC.                                   .                           Section.  Page. 

Definitions 1  1 

Notation  and  Numeration    ....'..  1  1 

Addition  .     ....    ^  '......  1  4 

Subtraction  .     ......     .     .     .     .  1  i) 

Multiplication    .     .     .     ....     .     .     .  1  11 

Division 1  16 

Cancelation  .     .     .     .     .     .     .  .-.     .     .     .  1  ,     It) 

Fractions .     .     .     . 1  22 

Decimals 1  35 

Symbols  of  Aggregation 1  49 

Percentage '  .     :  2  1 

Denominate  Numbers     .......  2  7 

Involution     .     .     . 2  22 

Evolution 2  25 

Ratio 2  42 

Proportion 2  46 

ELEMENTARY    ALGEBRA    AND     TRIGONOMETRIC 
FUNCTIONS. 

Elements  of  Algebra  ........  3  1 

Use  of  Letters .-...,  3  1 

Notation        .     .;.    .  "  .     |l|     .....  3  3 

Reading  Algebraic  Expressions    ....  3  6 

Positive  and  Negative  Quantities      ...  3  8 

Addition  and  Subtraction    .     .     c     .     .     .  3  10 

Symbols  of  Aggregation       .    '.     .     ...  3  17 

Multiplication    .     .     .....     ...  3.  It) 

v 


vi  CONTENTS. 

ELEMENTARY     ALGEBRA     AND    TRIGONOMETRIC 

FUNCTIONS — Continued.                                      Section.  Page. 

Division 3  25 

Factoring 3  28 

Fractions 3  34 

Mixed  Quantities  and  Complex  Fractions  .3  44 

Theory  of  Exponents 3  47 

Equations 3  51 

Trigonometric  Functions     ...  3  62 

Trigonometric  Tables 3  04 

Solution  of  Triangles 3  08 

Tables   of    Natural    Trigonometric    Func- 
tions     ............  3  75 

PHYSICS. 

Definitions 4  1 

Gravitation 4  0 

Tension  of  Gases 4  18 

'  Expansion  of  Gases 4  25 

Mixtures  of  Gases .-    .     .     .  4  35 

The  Air  Pump  .     .     .     .......  4  37 

Nature  of  Heat      .     ....     .     .     ....  4  41 

Temperature      ..........  4  42 

Expansion  of  Bodies  by  Heat 4  47 

Heat  Propagation 4  55 

Heat  Measurement      ........  4  57 

Specific  Heat     , 4  58 

Latent  Heat 4  04 

Sources  of  Heat 4  69 

Propagation  of  Light 4  72 

Reflection  and  Refraction    ...:..  4  74 

The  Spectrum .  4  83 

Double  Refraction  and  Polarization  ...  4  89 

Magnetism 4  96 

Electricity 4  101 

Electrostatics 4  102 

Electrodynamics 4  115 

Electromagnetism 4  122 


CONTENTS.  vii 

THEORETICAL  CHEMISTRY.  Section.     Page. 

Matter  and  Force .     5  1 

Weights  and  Measures    .......     5  8 

The  Balance '.....     5  13 

Laws  of  Chemical  Combination    ....     5  18 

Atomic  Weight:  Valence 5  21 

Symbols  and  Formulas 5  24 

The  Chemical  Elements .......     5  26 

Compound  Molecules       .......     5  35 

Volume  Relations  of  Molecules     .     ...     5  47 

Chemical  Calculations 5  55 

Stoichiometry 5  63 

Crystallography      .........     5  74 

Chemical  Operations  ........     5  81 

Acidimetry  and  Alkalimetry     .....     5  91 

Laboratory  Directions  and  Practical  Hints     5  1)6 

QUESTIONS  AND  EXAMPLES.  Section. 

Arithmetic,  Part  1 1 

Arithmetic,  Part  2 2 

Elementary  Algebra  and  Trigonometric  Functions  3 

Physics 4 

Theoretical  Chemistry .     .     .  5 


ARITHMETIC. 


DEFINITIONS. 

1.  Arithmetic  is  the  art  of  reckoning,  jos  i'bjeis^u'dy  of  '\  }'. 

numbers. 

*       '  *   '     j  }  j  ''>**'*»'*   • 

2.  A  unit  is  one,  or  a  single  thing,  as»a$?/  tone  bojv  She**  ' 

horse,  one  dozen. 

3.  A  number  is  a  unit  or  a  collection  of  units,  as  one, 
tliree  apples,  five  boys. 

4.  The  unit  of  a  number  is  one  of  the  collection  of 
units  which  constitutes  the   number.     Thus,   the   unit   of 
twelve  is  one,  of  twenty  dollars  is  one  dollar. 

5.  A  concrete  number  is  a  number  applied  to  some 
particular  kind  of  object  or  quantity,  as  three  horses,  five 
dollars,  ten  pounds. 

6.  An  abstract  number  is  a  number  that  is  not  applied 
to  any  object  or  quantity,  as  three,  five,  ten. 

7.  like  numbers  are  numbers  which  express  units  of 
the  same  kind,  as  six  days  and  ten  days,  two  feet  and  five  feet. 

8.  Unlike  numbers  are  numbers  which  express  units 
of  different  kinds,  as  ten  months  and  eight  miles,  seven  dol- 
lars and  five  feet. 

NOTATION  AND  NUMERATION. 

9.  Numbers  are  expressed  in  three  ways:    (1)  by  words; 
(2)  by  figures ;  (3)  by  letters. 

1C.     Notation  is  the  art  of  expressing  numbers  by  fig- 
ures or  letters. 

11.     Numeration   is  the  art  of  reading  the  numbers 
which  have  been  expressed  by  figures  or  letters. 


2  ARITHMETIC.  §  1 

12.  The  Arabic  notation  is  the  method  of  expressing 
numbers  by  figures.  This  method  employs  ten  different 
ligures  to  represent  numbers,  viz.  : 

Figures       0123456789 

Names  naught,  one    tivo  three  four  five     six  seven  eight  nine 
cipher, 
or  zero 

The  first  character  (0)  is  called  naught,  cipher,  or  zero, 

li^ii  feiiiidjng  alone  has  no  value. 
.The.  pjtiier.  nine-  ^figures  are  called  digits,  and  each  has  a 


Any  whole  number  is  called  an  integer. 

1  3.     As  there  are  only  ten  figures  used  in  expressing  num- 
bers, each  figure  must  have  a  different  value  at  different  times. 

14.  The  value  of  a  figure  depends  upon  its  position  in 
relation  to  others. 

15.  Figures   have  simple  values  and   local,  or  rela- 
tive, values. 

16.  The  simple  value  of  a  figure  is  the  value  it  ex- 
presses when  standing  alone. 

17.  The  local,  or  relative,  value  of  a  figure  is  the 
increased  value  it  expresses  by  having  other  figures  placed 
on  its  right. 

For  instance,   if  we  see  the  figure   6   standing 
alone,  thus  ......................  6 

we  consider  it  as  six  units,  or  simply  six. 

Place  another  6  to  the  left  of  it  ;  thus  ......  66 

The  original  figure  is  still  six  units,  but  the  sec- 
ond figure  is  ten  times  6,  or  6  tens. 

If  a  third  6  be  now  placed  still  one  place  further 
to  the  left,  it  is  .increased  in  value  ten  times  more, 
thus  making  it  6  hundreds  .........   «...  666 

A  fourth  6  would  be  6  thousands  ........         6666 

A  fifth  6  would  be  6   tens  of  thousands,  or 
sixty  thousands    ............   ......        66666 

A  sixth  6  would  be  6  hundreds  of  thousands  .     666666 
A  seventh  6  would  be  6  millions  .  .   6666666 


ARITHMETIC. 


The  entire  line  of  seven  figures  is  read  six  millions  six 
hundred  sixty-six  thousands  six  hundred  sixty-six. 

18.  The  increased  value  of  each  of  these  figures  is  its 
local,  or  relative,  value.     Each  figure  is  ten  times  greater  in 
value  than  the  one  immediately  on  its  right. 

19.  The  cipher  (0)  has  no  value  in  itself,  but  it  is  useful 
in  determining  the  place  of  other  figures.     To  represent  the 
number  four  hundred  five,  two  digits  only  are  necessary, 
one  to  represent  four  hundred,  and  the  other  to  represent 
five  units;  but  if  these  two  digits  are  placed  together,  as  45, 
the  4  (being  in  the  second  place)  will  mean  4  tens.     To  mean 
4  hundreds,  the  4  should  have  two  figures  on  its  right,  and  a 
cipher  is  therefore  inserted  in  the  place  usually  given  to  tens, 
to  show  that  the  number  is  composed  of  hundreds  and  units 
only,  and  that  there  are  no  tens.     Four  hundred  five  is  there- 
fore expressed  as  405.     If  the  number  were  four  thousand 
and  five,  two  ciphers  would  be  inserted;  thus,  4005.     If  it 
were  four  hundred  fifty,  it  would  have  the  cipher  at  the 
right-hand  side  to  show  that  there  were  no  units,  and  only 
hundreds  and  tens;    thus,   450.     Four  thousand  and  fifty 
would  be  expressed  4050,  the  first  cipher  indicating  that  there 
are  no  units  and  the  second  that  there  are  no  hundreds. 

20.  In  reading  numbers  that  have  been  represented  by 
figures,  it  is  usual  to  point  off  the  number  into  groups  of 
three  figures  each,  beginning  with  the  right-hand,  or  units, 
column,  a  comma  (,)  being  used  to  point  off  these  groups. 


Billions. 

Millions. 

Thousands. 

Units. 

rA 

d 

^ 

£3 

gj 

. 

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00 

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tc 

3 

d 

3 

go 

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PQ 

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Hundreds 

PQ 
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Billions. 

Hundreds 

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Millions. 

Hundreds 

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"S 

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0) 

Thousand; 

Hundreds 

Tens  of  U 

1 

4       3       2,1       9       8,7       6       5,4       3       2 

4  ARITHMETIC.  §  1 

In  pointing  off  these  figures,  begin  at  the  right-hand  figure 
and  count — units,  tens,  hundreds;  the  next  group  of  three 
figures  is  thousands;  therefore,  we  insert  a  comma  (,)  before 
beginning  with  them.  Beginning  at  the  figure  5,  we  say 
thousands^  tens  of  thousands,  hundreds  of  thousands,  and 
insert  another  comma.  We  next  read  millions,  tens  of  mil- 
lions, hundreds  of  millions  (insert  another  comma),  billions, 
tens  of  billions,  hundreds  of  billions. 

The  entire  line  of  figures  would  be  read:  four  hundred 
thirty-two  billions  one  hundred  ninety-eight  millions  seven 
hundred  sixty-five  thousands  four  hundred  thirty-two.  When 
we  thus  read  a  line  of  figures  it  is  called  numeration,  and 
if  the  "numeration  be  changed  back  to  figures,  it  is  called 
notation. 

For  instance,  the  writing  of  the  following  figures, 

72,584,623, 

would  be  the  notation,  and  the  numeration  would  be  sev- 
enty-two  millions  five  hundred  eighty-four  thousands  six  hun- 
dred twenty-three. 

21.  NOTE. — It  is   customary  to  leave  the  s  off  the  words  mil- 
lions, thousands,  etc.,  in  cases  like  the  above,  both  in  speaking  and 
writing;   hence,  the   above  would  usually  be  expressed  seventy -two 
million  five  hundred  eighty-four  thousand  six  hundred  twenty-three. 

22.  The  four  fundamental  processes  of  arithmetic  are 
addition,    subtraction,    multiplication,     and    division. 

They  are  called  fundamental  processes  because  all  operations 
in  arithmetic  are  based  upon  them. 


ADDITION. 

23.     Addition  is  fo&  process  of  finding  the  sum  of  two  or 

more  numbers.  The  sign  of  addition  is  -j- .  It  is  read  plus, 
and  means  more.  Thus,  5  -\-  6  is  read  5  plus  6,  and  means 
that  5  and  6  are  to  be  added. 

24-.  The  sign  of  equality  is  =  .  It  is  read  equals  or 
is  equal  to.  Thus,  5  +  6  —  11  may  be  read  5  plus  6 
equals  11. 


ARITHMETIC. 


5 


25.  Like  numbers  can  be  added,  but  unlike  numbers 
cannot  be  added.    Thus,  G  dollars  can  be  added  to  7  dollars, 
and  the  sum  will  be  13  dollars;   but  6  dollars  cannot  be 
added  to  7  feet. 

26.  The  following  table  gives  the  sum  of  any  two  num- 
bers from  1  to  12: 


1  and    1  is    2 

2  and    1  is    3 

3  and    1  is    4 

4  and    1  is    5 

1  and    2  is    3 

2  and    2  is    4 

3  and    2  is    5 

4  and    2  is    6 

1  and    3  is    4 

2  and    3  is    5 

3  and    3  is    6 

4  and    3  is    7 

1  and    4  is    5 

2  and    4  is    6 

3  and    4  is    7 

4  and    4  is    8 

1  and    5  is    6 

2  and    5  is    7 

3  and    5  is    8 

4  and    5  is    9 

1  and    6  is    7 

2  and    6  is    8 

3  and    6  is    9 

4  and    6  is  10 

1  and    7  is    8 

2  and    7  is    9 

3  and    7  is  10 

4  and    7  is  11 

1  and    8  is    9 

2  and    8  is  10 

3  and    8  is  11 

4  and    8  is  12 

1  and    9  is  10 

2  and    9  is  11 

3  and    9  is  12 

4  and    9  is  13 

1  and  10  is  11 

2  and  10  is  12 

3  and  10  is  13 

4  and  10  is  14 

1  and  11  is  12 

2  and  11  is  13 

3  and  11  is  14 

4  and  11  is  15 

1  and  12  is  13 

2  and  12  is  14 

3  and  12  is  15 

4  and  12  is  16 

5  and    1  is    6 

6  and    1  is    7 

7  and    1  is    8 

8  and    1  is    9 

Sand    2  is    7 

6  and    2  is    8 

7  and    2  is    9 

8  and    2  is  10 

5  and    3  is    8 

6  and    3  is    9 

7  and    3  is  10 

Sand    3  is"  11 

5  and    4  is    9 

6  and    4  is  10 

7  and    4  is  11 

8  and    4  is  12 

5  and    5  is  10 

6  and    5  is  11 

7  and    5  is  12 

8  and    5  is  13 

5  and    6  is  11 

6  and    6  is  12 

7  and    6  is  13 

8  and    6  is  14 

5  and    7  is  12 

6  and    7  is  13 

7  and    7  is  14 

8  and    7  is  15 

5  and    8  is  13 

6  and    8  is  14 

7  and    8  is  15 

8  and    8  is  16 

5  and    9  is  14 

6  and    9  is  15 

7  and    9  is  16 

8  and    9  is  17 

5  and  10  is  15 

6  and  10  is  16 

7  and  10  is  17 

8  and  10  is  18 

5  and  11  is  16 

6  and  11  is  17 

7  and  11  is  18 

8  and  11  is  19 

5  and  12  is  17 

6  and  12  is  18 

7  and  12  is  19 

8  and  12  is  20 

9  and    1  is  10 

10  and    1  is  11 

11  and    1  is  12 

12  and    1  is  13 

9  and    2  is  11 

10  and    2  is  12 

11  and    2  is  13 

12  and    2  is  14 

9  and    3  is  12 

10  and    3  is  13 

11  and    3  is  14 

12  and    3  is  15 

9  and    4  is  13 

10  and    4  is  14 

11  and    4  is  15 

12  and    4  is  16 

9  and    5  is  14 

10  and    5  is  15 

11  and    5  is  16 

12  and    5  is  17 

9  and    6  is  15 

10  and    6  is  16 

11  and    6  is  17 

12  and    6  is  18 

9  and    7  is  16 

10  and    7  is  17 

11  and    7  is  18 

12  and    7  is  19 

9  and    8  is  17 

10  and    8  is  18 

11  and    8  is  19 

12  and    8  is  20 

9  and    9  is  18 

10  and    9  is  19 

11  and    9  is  20 

12  and    9  is  21 

9  and  10  is  19 

10  and  10  is  20 

11  and  10  is  21 

12  and  10  is  22 

9  and  11  is  20 

10  and  11  is  21 

11  and  11  is  22 

12  and  11  is  23 

9  and  12  is  21 

10  and  12  is  22 

11  and  12  is  23 

12  and  12  is  24 

This  table  should  be  carefully  committed  to  memory.  Since  0  has  no 
value,  the  sum  of  any  number  and  0  is  the  number  itself ;  thus  17  and  0 
is  17. 

27.  For  addition,  place  the  numbers  to  be  added  directly 
under  each  other,  taking  care  to  place  units  under  units,  tens 
under  tens,  hundreds  under  hundreds,  and  so  on. 


6  ARITHMETIC.  §  1 

When  the  numbers  are  thus  written,  the  right-hand  figure 
of  one  number  is  placed  directly  under  the  right-hand  figure 
of  the  one  above  it,  thus  bringing"  units  under  units,  tens 
under  tens,  etc.  Proceed  as  in  the  following  examples : 

28.  EXAMPLE.— What  is  the  sum  of  131,  222,  21,  2,  and  413  ? 

SOLUTION. —  131 

222 

2  1 

2 

413 
sum    789    Ans. 

EXPLANATION.  —  After  placing  the  numbers  in  proper 
order,  begin  at  the  bottom  of  the  right-hand,  or  imits,  col- 
umn, and  add,  mentally  repeating  the  different  sums.  Thus, 
three  and  two  are  five  and  one  are  six  and  two  are  eight  and 
one  are  nine,  the  sum  of  the  numbers  in  units  column. 
Place  the  9  directly  beneath  as  the  first,  or  units,  figure  in 
the  sum. 

The  sum  of  the  numbers  in  the  next,  or  tens,  column 
equals  8  tens,  which  is  the  second,  or  tens,  figure  in  the 
sum. 

The  sum  of  the  numbers  in  the  next,  or  hundreds,  column 
equals  7  hundreds,  which  is  the  third,  or  hundreds,  figure  in 
the  sum. 

The  sum,  or  answer,  is  789. 

29.  EXAMPLE.— What  is  the  sum  of  425,  36,  9,215,  4,  and  907  ? 

SOLUTION. —  425 

36 

9215 
4 

907 
27 
60 

1500 
9000 


sum    10587    Ans. 
EXPLANATION. — The  sum  of  the  numbers  in  the  first,  or 


§  1  ARITHMETIC.  7 

units,  column  is  seven  and  four  are  eleven  and  five  are  six- 
teen and  six  are  twenty-two  and  five  are  twenty-seven,  or 
27  units;  i.  e.,  two  tens  and  seven  units.  Write  27  as 
shown.  The  sum  of  the  numbers  in  the  second,  or  tens, 
column  is  six  tens,  or  60.  Write  60  underneath  27,  as 
shown.  The  sum  of  the  numbers  in  the  third,  or  hundreds, 
column  is  15  hundreds,  or  1,500.  Write  1,500  under  the  two 
preceding  results  as  shown.  There  is  only  one  number  in 
the  fourth,  or  thousands,  column,  9,  which  represents  9,000. 
Write  9,000  under  the  three  preceding  results.  Adding 
these  four  results,  the  sum  is  10,587,  which  is  the  sum  of 
425,  36, '9,215,  4,  and  907. 

NOTE. — It  frequently  happens  when  adding  a  long  column  of  fig- 
ures, that  the  sum  of  two  numbers,  one  of  which  does  not  occur  in  the 
addition  table,  is  required.  Thus,  in  the  first  column  above,  the  sum 
of  16  and  6  was  required.  We  know  from  the  table  that  6  +  6  =  12; 
hence,  the  first  figure  of  the  sum  is  2.  Now,  the  sum  of  any  number 
less  than  20  and  of  any  number  less  than  10  must  be  less  than  30,  since 
20  +  10  =  30 ;  therefore,  the  sum  is  22.  Consequently,  in  cases  of  this 
kind,  add  the  first  figure  of  the  larger  number  to  the  smaller  number, 
and  if  the  result  is  greater  than  9,  increase  the  second  figure  of  the  larger 
number  by  1.  Thus,  44  +  7=?  4  +  7  =  11;  hence,  44  +  7  =  51. 

3O.     The  addition  may  also  be  performed  as  follows: 

425 
36 

9215 
4 

907 
sum  10587  Ans. 

EXPLANATION. — The  sum  of  the  numbers  in  units  column 
is  27  units,  or  2  tens  and  7  units.  Write  the  7  units  as  the 
first,  or  right-hand,  figure  in  the  sum.  Reserve  the  two 
tens  and  add  them  to  the  figures  in  tens  column.  The  sum 
of  the  figures  in  the  tens  column,  plus  the  2  tens  reserved 
and  carried  from  the  units  column,  is  8,  which  is  written 
down  as  the  second  figure  in  the  sum.  There  is  nothing  to 
carry  to  the  next  column,  because  8  is  less  than  10.  The 
sum  of  the  numbers  in  the  next  column  is  15  hundreds,  or 
1  thousand  and  5  hundreds.  Write  down  the  5  as  the  third, 
or  hundreds,  figure  in  the  sum  and  carry  the  1  to  the  next 

1-2 


8  ARITHMETIC.  §  1 

column.     1  +  9  =  10,  which  is  written  down  at  the  left  of 
the  other  figures. 

The  second  method  saves  space  and  figures,  but  the  first 
is  to  be  preferred  when  adding  a  long  column. 

31.  EXAMPLE. — Add  the  numbers  in  the  column  below: 

SOLUTION.—  890 

82 

90 
393 
281 

80 
770 

83 
492 

80 
383 

84 
191 
sum  3899  Ans. 

EXPLANATION. — The  sum  of  the  digits  in  the  first  column 
equals  19  units,  or  1  ten  and  9  units.  Write  down  the  9  and 
carry  1  to  the  next  column.  The  sum  of  the  digits  in  the 
second  column  4- 1  is  109  tens,  or  10  hundreds  and  9  tens. 
Write  down  the  9  and  carry  the  10  to  the  next  column. 
The  sum  of  the  digits  in  this  column  plus  the  10  reserved 
is  38. 

The  entire  sum  is  3,899. 

32.  Rule. — I.     Begin  at  the  right,  add  each  column  sep- 
arately, and  write  the  sum,  if  it  be  only  one  figure,  under  the 
column  added. 

II.  If  the  sum  of  any  column  consists  of  two  or  more 
figures,  put  the  right-hand  figure  of  the  sum  under  that 
column  and  add  the  remaining  figure  or  figures  to  the  next 
column. 

33.  Proof. —  To  prove  addition,  add  each  column  from 
top  to  bottom.     If  you  obtain  the  same  result  as  by  adding 
from  bottom  to  top,  the  work  is  probably  correct. 


gl  ARITHMETIC. 

EXAMPLES    FOR    PRACTICE. 

34.   Find  the  sum  of: 

(a)  104  +  203  +  613  +  214. 

(b)  1 , 875  +  3, 143  +  5, 826  +  10, 832. 

(c)  4,865  +  2,145  +  8,173  +  40,084. 

(d)  14,204  +  8,173  +  1,065  +  10,042.     . 

(e)  10,832  +  4,145  +  3,133  +  5,872. 
(/)  214  +  1,231  +  141  +  5,000. 

Or)  123  +  104  +  425  +  126  +  327. 

(//)     6,354  +  2,145  +  2,042  +  1,111  +  3,333. 


(a)  1,134. 

(£)  21,676. 

(0  55,267. 

(d)  33,484. 

(e)  23,982. 
(/)  6,586. 
(-)  1,105. 
(//)  14,985. 


SUBTRACTION. 

35.  In  arithmetic,  subtraction  is  the  process  of  finding- 
how  much  greater  one  number  is  than  another. 

The  greater  of  the  two  numbers  is  called  the  minuend. 
The  smaller  of  the  two  numbers  is  called  the  subtrahend. 
The  number  left  after  subtracting  the  subtrahend  from 
the  minuend  is  called  the  difference,  or  remainder. 

36.  The  sign  of   subtraction  is  — .     It  is  read  minus, 
and  means  less.     Thus,  12  --  7  is  read  12  minus  7,  and  means 
that  7  is  to  be  taken  from  12. 

37.  EXAMPLE.— From  7,568  take  3,425. 
SOLUTION. —  minuend    7568 

subtrahend    3425 

remainder    4143    Ans. 

EXPLANATION. — Begin  at  the  right-hand,  or  units,  column 
and  subtract  in  succession  each  figure  in  the  subtrahend  from 
the  one  directly  above  it  in  the  minuend,  and  write  the  remain- 
ders below  the- line.  The  result  is  the  entire  remainder. 

38.  When  there  are  more  figures  in  the  minuend  than 
in  the  subtrahend,  and  when  some  figures  in  the  minuend 
are  less  than  the  figures  directly  under  them  in  the  subtra- 
hend, proceed  as  in  the  following  example : 

EXAMPLE.— From  8,453  take  844. 
SOLUTION. —  minuend    8453 

subtrahend       844 

remainder    7609    Ans. 


10  ARITHMETIC.  §  I 

EXPLANATION. — Begin  at  the  right-hand,  or  units,  column 
to  subtract.  We  cannot  take  4  from  3,  and  must,  therefore, 
borrow  1  from  5  in  tens  column  and  annex  it  to  the  3  in 
units  column.  The  1  ten  =  10  units,  which  added  to  the  3 
in  units  column  =  13  units.  4  from  13  =  9,  the  first,  or 
units,  figure  in  the  remainder. 

Since  we  borrowed  1  from  the  5,  only  4  remains ;  4  from 
4  =  0,  the  second,  or  tens,  figure.  We  cannot  take  8  from 
4,  and  must,  therefore,  borrow  1  from  8  in  thousands  column. 
Since  1  thousand  =  10  hundreds,  10  hundreds +  4  hundreds 
=  14  himdreds,  and  8  from  14  ='C,  the  third,  or  hundreds, 
figure  in  the  remainder. 

Since  we  borrowed  1  from  8,  only  7  remains,  from  which 
there  is  nothing  to  subtract ;  therefore,  7  is  the  next  figure 
in  the  remainder,  or  answer. 

The  operation  of  borrowing  is  performed  by  mentally 
placing  1  before  the  figure  following  the  one  from  which  it 
is  borrowed.  In  the  above  example  the  1  borrowed  from  5 
is  placed  before  3,  making  it  13,  from  which  we  subtract  4. 
The  1  borrowed  from  8  is  placed  before  4,  making  14,  from 
which  8  is  taken. 

39.  EXAMPLE.— Find  the  difference  between  10,000  and  8,763. 

SOLUTION. —  minuend  10000 
subtrahend  8763 
remainder  1237  Ans. 

EXPLANATION. — In  the  above  example  we  borrow  1  from 
the  second  column  and  place  it  before  0,  making  10;  3 
from  10  =  7.  In  the  same  way  we  borrow  1  and  place  it 
before  the  next  cipher,  making  10 ;  but  as  we 'have  borrowed 
1  from  this  column  and  have  taken  it  to  the  units  column, 
only  0  remains  from  which  to  subtract  6 ;  6  from  9  =  3. 
For  the  same  reason  we  subtract  7  from  9  and  8  from  9  for 
the  next  two  figures,  and  obtain  a  total  remainder  of  1,237. 

40.  Rule. — Place   the  subtrahend  (or  smaller  number] 
under  the  minuend  (or  larger  number),  in  the  same  manner  as 
for  addition,  and  proceed  as  in  Arts.  37,  38,  and  39. 


ARITHMETIC. 


11 


41.  Proof. —  To  prove  an  example  in  subtraction,  add 
the  subtrahend  and  the  remainder.  The  sum  should  equal 
the  minuend.  If  it  does  not,  a  mistake  has  been  made,  and 
the  work  should  be  done  over. 

Proof  of  the  above  example : 

subtrahend       8763 

remainder       1237 

minuend   10000 


42. 


w 

(/) 
(*•) 
(*) 


EXAMPLES  FOR  PRACTICE. 

From: 

94,278  take  62,574. 
.53,714  take  25,824. 
71,832  take  58,109. 
20,804  take  10,408.  . 

310,465  take  102,141. 
(81,043  +  1,041)  take  14,831. 
(20,482  +  18,216)  take  21,214. 
(2,040  +  1,213  +  542)  take  3,791. 


(d) 

w 

(/") 

Or) 


31,704. 

27,890. 

13,723. 

10,396. 

208,324. 

67,253. 

17,484. 

4. 


MULTIPLICATION. 

43.  To  multiply  a  number  is  to  add  it  to  itself  a  certain 
number  of  times. 

44.  Multiplication  is  the  process  of  multiplying  one 
mimber  by  another. 

The  number  thus  added  to  itself,  or  the  number  to  be 
multiplied,  is  called  the  multiplicand. 

The  number  which  shows  how  many  times  the  multi- 
plicand is  to  be  taken,  or  the  number  by  which  we  multiply, 
is  called  the  multiplier. 

The  result  obtained  by  multiplying  is  called  the  product. 

45.  The  sign  of  multiplication  is  X .     It  is  read  times  or 
multiplied  by.     Thus,  9  X  6  is  read  9  times  6,  or  9  multiplied 
by  6. 

46.  It  matters  not  in  what  order  the  numbers  to  be  multi- 
plied together  are  placed.     Thus,  6  X  9  is  the  same  as  9  X  6. 


ARITHMETIC. 


. 


In  the  following  table,  the  product  of  any  two  num- 


bers  (neither  of  which  exceeds  12)  may  be  found  : 


1  times    1  is      1 

2  times    1  is      2 

3  times    1  is      3 

1  times    2  is      2 

2  times    2  is      4 

3  times    2  is      6 

1  times    3  is      3 

2  times    3  is      6 

3  times    3  is      9 

1  times    4  is      4 

2  times    4  is      8 

3  times    4  is    12 

1  times    5  is      5 

2  times    5  is    10 

3  times    5  is    15 

1  times    6  is      6 

2  times    6  is    12 

3  times    6  is    18 

1  times    7  is      7 

2  times    7  is    14 

3  times    7  is    21 

1  times    8  is      8 

2  times    8  is    16 

3  times    8  is    24 

1  times    9  is      9 

2  times    9  is    18 

3  times    9  is    27 

1  times  10  is    10 

2  times  10  is    20 

3  times  10  is    30 

1  times  11  is    11 

2  times  11  is    22 

3  times  11  is    33 

1  times  12  is    12 

2  times  12  is    24 

3  times  12  is    36 

4  times    1  is      4 

5  times    1  is      5 

6  times    1  is      6 

4  times    2  is      8 

5  times    2  is    10 

6  times    2  is    12 

4  times    3  is    12 

5  times    3  is    15 

6  times    3  is    18 

4  times    4  is    16 

5  times    4  is    20 

6  times    4  is    24 

4  times    5  is    20 

5  times    5  is    25 

6  times    5  is    30 

4  times    6  is    24 

5  times    6  is    30 

6  times    6  is    36 

4  times    7  is    28 

5  times    7  is    35 

6  times    7  is    42 

4  times    Sis    32 

5  times    8  is    40 

6  times    8  is    48 

4  times    9  is    36  • 

5  times    9  is    45 

6  times    9  is    54 

4  times  10  is    40 

5  times  10  is    50 

6  times  10  is    60 

4  times  11  is    44 

5  times  11  is    55 

6  times  11  is    66 

4  times  12  is    48 

5  times  12  is    60 

6  times  12  is    72 

7  times    1  is      7 

8  times    1  is      8 

9  times    1  is      9 

7  times    2  is    14 

8  times    2  is    16 

9  times    2  is    18 

7  times    3  is    21 

.  8  times    3  is    24 

9  times    3  is    27 

7  times    4  is    28 

8  times    4  is    32 

9  times    4  is    36 

7  times    5  is    35 

8  times    5  is    40 

9  times    5  is    45 

7  times    6  is    42 

8  times    6  is    48 

9  times    6  is    54 

7  times    7  is    49 

8  times    7  is    56 

9  times    7  is    63 

7  times    8  is    56 

8  times    8  is    64 

9  times    8  is    72 

7  times    9  is    63 

8  times    9  is    72 

9  times    9  is    81 

7  times  10  is    70 

8  times  10  is    80 

9  times  10  is    90 

7  times  11  is    77 

8  times  11  is    88 

9  times  11  is    99 

7  times  12  is    84 

8  times  12  is    96 

9  times  12  is  108 

10  times    1  is    10 

11  times    1  is    11 

12  times    1  is    12 

10  times    2  is    20 

11  times    2  is    22 

12  times    2  is    24 

10  times    3  is    30 

11  times    3  is    33 

12  times    3  is    36 

10  times    4  is    40 

11  times    4  is    44 

12  times    4  is    48 

10  times    5  is    50 

11  times    5  is    55 

12  times    5  is    60 

10  times    6  is.  60 

11  times    6  is    66 

12  times    6  is    72 

10  times    7  is    70 

11  times    7  is    77 

12  times    7  is    84 

10  times    8  is    80 

11  times    8  is    88 

12  times    8  is    96 

10  times    9  is    90 

11  times    9  is    99 

12  times    9  is  108 

10  times  10  is  100 

11  times  10  is  110 

12  times  10  is  120 

10  times  11  is  110 

11  times  11  is  121 

12  times  11  is  132 

10  times  12  is  120 

11  times  12  is  132 

12  times  12  is  144 

This  table  should  be  carefully  committed  to  memory. 
Since  0  has  no  value,  the  product  of  0  and  any  number  is  0. 


§  1  ARITHMETIC.  13 

48.     To  multiply  a  number  by  one  iigure  only  : 

EXAMPLE. — Multiply  425  by  5. 
SOLUTION. —        multiplicand        425 
multiplier  5 

product     2125    Ans. 

EXPLANATION. — For  convenience,  the  multiplier  is  gener- 
ally written  under  the  right-hand  figure  of  the  multiplicand. 
On  looking  in  the  multiplication  table,  we  see  that  5x5  are 
25.  Multiplying  the  first  figure  at  the  right  of  the  multi- 
plicand, or  5,  by  the  multiplier,  5,  it  is  seen  that  5  times  5 
units  are  25  units,  or  2  tens  and  5  units.  Write  the  5  units 
in  units  place  in  the  product,  and  reserve  the  2  tens  to  add 
to  the  product  of  tens.  Looking  in  the  multiplication  table 
again,  we  see  that  5x2  are  10.  Multiplying  the  second 
figure  of  the  multiplicand  by  the  multiplier,  5,  we  see  that 
5  times  2  tens  are  10  tens,  and  10  tens  plus  the  2  tens  reserved 
are  12  tens,  or  1  hundred  plus  2  tens.  Write  the  2  tens  in 
tens  place,  and  reserve  the  1  hundred  to  add  to  the  product 
of  hundreds.  Again,  we  see  by  the  multiplication  table  that 
5x4  are  20.  Multiplying  the  third,  or  last,  figure  of  the 
multiplicand  by  the  multiplier,  5,  we  see  that  5  times  4  hun- 
dreds are  20  hundreds,  and  20  hundreds  plus  the  1  hundred  re- 
served are  21  hundreds,  or  2  thousands  and  1  hundred,  which 
we  write  in  thousands  and  hundreds  places,  respectively. 

Hence,  the  product  is  2,125. 

This  result  is  the  same  as  adding  425  five  times.     Thus, 

425 
425 
425 
425 

425 
sum  2125  Ans. 


EXAMPLES  FOB  PRACTICE. 

49.      Find  the  product  of  : 


(a)  61,483  X  6. 

(&)  12,375  X  5. 

(c)  10,426  X  7. 

(d)  10,835X3. 


Ans. 


(a)  368,898. 

(£)  61,875. 

(c)  72,982. 

(d)  32,505. 


14  ARITHMETIC. 


(<?)  98,376X4. 

(/)  10,873X8. 

(g)  71,543X9. 

(h)  218,734  X  2. 


Ans.  - 


(<?)  393,504. 

(/)  86,984. 

(g)  643,887. 

(//)  437,468. 


50.  To  multiply  a  number  by  two  or  more  figures : 

EXAMPLE.— Multiply  475  by  234. 

SOLUTION. —     multiplicand  475 

multiplier  234 

1900 
1425 
950 
product    111150    Ans. 

EXPLANATION. — For  convenience,  the  multiplier  is  gener- 
ally written  under  the  multiplicand,  placing  units  under 
units,  tens  under  tens,  etc. 

We  cannot  multiply  by  234  at  one  operation;  we  must, 
therefore,  multiply  by  the  parts  and  then  add  the  partial 
products. 

The  parts  by  which  we  are  to  multiply  are  4  units,  3  tens, 
and  2  hundreds.  4  times  475  =  1,900,  the  first  partial 
product;  3  times  475  =  1,425,  the  second  partial  product, 
the  right-hand  figure  of  which  is  written  directly  under 
the  figure  multiplied  by,  or  3 ;  2  times  475  =  950,  the  third 
partial  product,  the  right-hand  figure  of  which  is  written 
directly  under  the  figure  multiplied  by,  or  2. 

The  sum  of  these  three  partial  products  is  111,150,  which 
is  the  entire  product. 

51.  Rule. — I.      Write  the  multiplier  under  the  multipli- 
cand, so  that  units  are  under  units,  tens  under  tens,  etc. 

II.  Begin  at  the  right  and  multiply  each  figure  of  the 
multiplicand  by   each    successive  figure    of  the    multiplier, 
placing  the  right-hand  figure  of  each  partial  product  directly 
under  the  figure  used  as  a  multiplier. 

III.  The  sum   of  the  partial  products   ivill  equal  the 
required  product. 


ARITHMETIC. 


15 


52.  Proof. — Review  the  ivork  carefully,  or  multiply  the 
multiplier  by  the  multiplicand;  if  the  results  agree,  the  work 
is  correct. 

53.  When  there  is  a  cipher  in  the  multiplier,  multiply 
by  it  the  same  as  with  the  other  figures.     Thus, 

(a)  (6)  (O        (</) 

0  2  15        708 

X—  X_?.  X  0      X 0 

0  Ans.  0  Ans.  0  0  Ans.    000  Ans. 


8114 

203 

9342 

0000 
6228 
632142  Ans. 


4008 
305 
20040 
0000 
12024 
1222440  Ans. 


tr) 

31264 
1002 
62528 
00000 
00000 
31264 
31326528  Ans. 


When  multiplying  by  a  number  containing  a  cipher,  the 
work  may  be  shortened  by  writing  the  first  cipher  of  the 
partial  product,  then  multiplying  by  the  next  figure  of 
the  multiplier  and  writing  the  partial  product  alongside  of 
the  cipher.  Thus,  examples  (e)  and  (g)  above  might  have 
been  solved  in  the  following  manner: 

3114  31264 

203  1002 


9342 
62280 
632142  Ans. 


62528 
3126400 
31326528  Ans. 


EXAMPLES    FOR    PRACTICE. 


54. 

(*) 

(*) 

(d) 
W 

U-) 

(*) 


Find  the  product  of: 
3,842  X  26. 
3,716  X  45. 
1,817X124. 
675  X  38. 
1,875  X  33. 
4,836  X  47. 
5,682  X  543. 
3,257  X  246. 


Ans. 


(a) 

(*) 
(c) 
(</) 

w 

</) 

Or) 
(*) 


99,892. 

167,220. 

225,308. 

25,650. 

61,875. 

227,292. 

3,085,326. 

801,222. 


16  ARITHMETIC. 


(/)  2,875  X  302. 

(/)  17,819  X  1,004. 

(£)  38,674  X  205. 

(/)  18,304X100.  Ans. 

(///)  7,834X10. 

(«)  87,543  X  1,000. 

(o)  48,768  X  100. 


(/)  868,250. 

(/)  17,890,276. 

(k)  7,928,170. 

(/)  1,830,400. 

(m)  78,340 

(it)  87,543,000. 

(o)  4,876,300. 


DIVISION. 

55.  Division  is  the  process  of  finding  how  many  times 
one  number  is  contained  in  another  of  the  same  kind. 

The  number  to  be  divided  is  called  the  dividend. 
The  number  by  which  we  divide  is  called  the  divisor. 
The  number  which  shows  how  many  times  the  divisor  is 
contained  in  the  dividend  is  called  the  quotient. 

56.  The  sign  of  division  is  -K     It  is  read  divided  by. 
54  -r-  9  is  read  54  divided  by  9.      Another  way  to  write  54 

divided  by  9  is-^-.     Thus,  54  -f-  9  =  6,  or  y  =  G. 

In  both  of  these  cases,  54  is  the  dividend  and  9  is  the 
divisor. 

Division  is  the  reverse  of  multiplication. 

57.  To  divide  when  the  divisor  consists  of  but  one 
figure,  proceed  as  in  the  following  example  : 

EXAMPLE.—  What  is  the  quotient  of  875  -*-  7  ? 

divisor  dividend  quotient 
SOLUTION.—  7)875(125    Ans. 

_7_ 

17 

14 


35 
JJJ> 

remainder        0 

EXPLANATION.—  7  is  contained  in  8  hundreds,  1  hundred 
times.  Place  the  1  as  the  first,  or  left-hand,  figure  of  the 
quotient.  Multiply  the  divisor,  7,  by  the  1  hundred  of  the 


§  1  ARITHMETIC.  17 

quotient,  and  place  the  product,  7  hundreds,  under  the  8 
hundreds  in  the  dividend,  and  subtract.  Beside  the  re- 
mainder, 1,  bring  down  the  next,  or  tens,  figure  of  the 
dividend,  in  this  case  7,  making  17  tens;  7  is  contained  in 
17,  2  times.  Write  the  2  as  the  second  figure  of  the  quo- 
tient. Multiply  the  divisor,  7,  by  the  2  in  the  quotient,  and 
subtract  the  product  from  17.  Beside  the  remainder,  3, 
bring  down  the  units  figure  of  the  dividend,  making  35  units. 
7  is  contained  in  35,  5  times,  which  is  placed  in  the  quotient. 
Multiplying  the  divisor  by  the  last  figure  of  the  quotient, 
5  times  7  =  35,  which  subtracted  from  35,  under  which  it 
is  placed,  leaves  0.  Therefore,  the  quotient  is  125.  This 
method  is  called  long  division. 

58.  In  short  division,  only  the  divisor,  dividend,  and 
quotient  are  written,  the  operations  being  performed  men- 
tally. 

dividend 
divisor  7  )81735 

quotient         125    Ans. 

The  mental  operation  is  as  follows:  7  is  contained  in  8, 
once  and  1  remainder;  imagine  1  to  be  placed  before  7, 
making  17;  7  is  contained  in  17,  2  times  and  3  over;  imag- 
ine 3  to  be  placed  before  5,  making  35 ;  7  is  contained  in  35, 
5  times.  These  partial  quotients,  placed  in  order  as  they 
are  found,  make  the  entire  quotient  125. 

59.  If  the  divisor  consists  of  two  or  more  figures,  pro- 
ceed as  in  the  following  example : 

EXAMPLE.— Divide  2,702,826  by  63. 

divisor         dividend  qttotient 

SOLUTION.—  63)2702826(42902    Ans. 

252 


18  ARITHMETIC.  §  1 

EXPLANATION. — As  63  is  not  contained  in  the  first  two 
figures,  27,  we  must  use  the  first  three  figures,  270.  Now, 
by  trial  we  must  find  how  many  times  63  is  contained  in  270. 
6  is  contained  in  the  first  two  figures  of  270,  4  times.  Place  the 
4  as  the  first,  or  left-hand,  figure  in  the  quotient.  Multiply  the 
divisor,  63,  by  4,  and  subtract  the  product,  252,  from  270. 
The  remainder  is  18,  beside  which  we  write  the  next  figure 
of  the  dividend,  2,  making  182.  Now,  6  is  contained  in  the 
first  two  figures  of  182,  3  times,  but  on  multiplying  63  by  3, 
we  see  that  the  product,  189,  is  too  great,  so  we  try  2  as 
the  second  figure  of  the  quotient.  Multiplying  the  divi- 
sor, 63,  by  2  and  subtracting  the  product,  126,  from  182,  the 
remainder  is  56,  beside  which  we  bring  down  the  next  figure 
of  the  dividend,  making  568.  6  is  contained  in  56  about  9 
times.  Multiply  the  divisor,  63,  by  9  and  subtract  the  prod- 
uct, 567,  from  568.  The  remainder  is  1,  and  bringing 
down  the  next  figure  of  the  dividend,  2,  gives  12.  As  12  is 
smaller  than  63,  we  write  0  in  the  quotient  and  bring  down 
the  next  figure,  6,  making  126.  63  is  contained  in  126,  2 
times,  without  a  remainder.  Therefore,  42,902  is  the 
quotient. 

6O.  Rule. — I.  Write  the  divisor  at  the  left  of  the  divi- 
dend^ ivith  a  line  between  them. 

II.  Find-  how  many  times  the  divisor  is  contained  in  the 
lowest  number  of  the  left-hand  figures  of  the  dividend  that 
will  contain  it,  and  write  the  result  at   the   right  of  the 
dividend,   with  a  line  between,  for  the  first  figure  of  the 
quotient. 

III.  Multiply  the  divisor  by  this  quotient;  write  the  prod- 
uct under  the  partial  dividend  used,  and  subtract,  annexing 
to  the  remainder  the  next  figure  of  the  dividend.     Divide  as 
before,  and  thus  continue  until  all  the  figures  of  the  dividend 
have  been  used. 

IV.  If  any  partial  dividend  will  not  contain  the  divisor, 
write  a  cipher  in  the  quotient,  annex  the  next  figure  of  the 
dividend,  and  proceed  as  before. 


ARITHMETIC. 


19 


V.  If  there  be  at  last  a  remainder,  write  it  after  the 
quotient,  ivith  the  divisor  underneath. 

61.  Proof. — Multiply  the  quotient  by  the  divisor  and  add 
the  remainder,  if  there  be  any,  to  the  product.  The  result 
ivill  be  the  dividend.  Thus, 

divisor    dividend    quotient 
63  )  4235  (  67£|    Ans. 
378 


PROOF. — 


remainder 

quotient 

divisor 


remainder 
dividend 


455 

441 

14 

67 

63 

201 

402 

4221 

14 

4235 


EXAMPLES   FOB  PRACTICE. 

63. 

Divide  the  following: 

(a) 

126,498  by  58. 

(a) 

2,181. 

(*) 

3,207,594  by  767. 

w 

4,182. 

to 

11,408,202  by  234. 

to 

^8,753. 

(d) 

2,100,315  by  581.                                  . 

(</) 

3,615. 

to 

969,936  by  4,008. 

to 

242. 

(/) 

7,481,888  by  1,021. 

(/) 

7,328. 

fc) 

1,525,915  by  5,003.       .' 

Cirt 

305. 

w 

1,646,301  by  381. 

» 

4,321. 

CA^CELATKXtf. 

63.  Cancelation   is  the  process  of  shortening"  opera- 
tions in  division  by  casting  out  equal  factors  from  both 
dividend  and  divisor. 

64.  The  factors  of  a  number  are  those  numbers,  which, 
when  multiplied  together,  produce  the  given  number.    Thus, 
5  and  3  are  the  factors  of  15,  since  5x3  =  15.     Likewise, 
8  and  7  are  the  factors  of  56,  since  8x7  =  56. 


20  ARITHMETIC.  §  1 

65.  A  prime  number  is  one  which  cannot  be  divided 
by  any  number  except  itself  and  1.     Thus,  2,  3,  11,  29,  etc. 
are  prime  numbers. 

66.  A   prime   factor  is   any  factor   that   is   a   prime 
number. 

Any  number  that  is  not  a  prime  is  called  a  composite 
number,  and  may  be  produced  by  multiplying  together  its 
prime  factors.  Thus,  60  is  a  composite  number,  and  is 
equal  to  the  product  of  its  prime  factors,  2x2x3x5. 

67.  Canceling  equal  factors  from    both   dividend  and 
divisor  does  not  change  the  quotient. 

The  canceling  of  a  factor  in  both  dividend  and  divisor  is 
the  same  as  dividing  them  both  by  the  same  number,  and 
this,  evidently,  does  not  change  the  quotient. 

Write  the  numbers  forming  the  dividend  above  a  hori- 
zontal line,  and  those  forming  the  divisor  below  it;  then 
cancel  the  equal  factors. 

68.  EXAMPLE.— Divide  4  X  45  X  60  by  9  X  24. 

SOLUTION. — Placing  the  dividend  over  the  divisor,  and  canceling, 
5       10 

£x*xff  =  «U*  Ans. 

% 

EXPLANATION. — The  4  in  the  dividend  and  the  24  in  the 
divisor  are  both  divisible  by  4,  since  4  divided  by  4  equals  1 , 
and  24  divided  by  4  equals  6.  Cross  off  the  4  and  write  the 
1  over  it ;  also,  cross  off  the  24  and  write  the  6  under  it.  Thus, 

1 
£  X  45  X  60  _ 

9X2* 

6 

GO  in  the  dividend  and  6  in  the  divisor  are  divisible  by  0, 
since  GO  divided  by  6  equals  10,  and  G  divided  by  G  equals  1. 
Cross  off  the  60  and  write  10  over  it;  also,  cross  off  the  G 
and  write  1  under  it.  Thus, 

1  10 

*  X  45  X  W 


§  1  ARITHMETIC.  21 

Again,  45  in  the  dividend  and  9  in  the  divisor  are  divisi- 
ble by  9,  since  45  divided  by  9  equals  5,  and  9  divided  by  9 
equals  1.  Cross  off  the  45  and  write  the  5  over  it;  also, 
cross  off  the  9  and  write  the  1  under  it.  Thus, 

1      5      10 


9X& 

1      $ 

Since  there  are  no  two  remaining  numbers  (one  in  the 
dividend  and  one  in  the  divisor)  divisible  by  any  number 
except  1,  without  a  remainder,  it  is  impossible  to  cancel 
further. 

Multiply  all  the  uncanceled  numbers  in  the  dividend 
together  and  divide  their  product  by  the  product  of  all  the 
uncanceled  numbers  in  the  divisor.  The  result  will  be  the 
quotient.  The  product  of  all  the  uncanceled  numbers  in 
the  dividend  equals  5xlXlO  —  50;  the  product  of  all  the 
uncanceled  numbers  in  the  divisor  equals  1x1  =  1. 

1      5      10 

Hence,  lXf^  =  lXaXlO  =  50.Ans. 

1  % 

69.  Rule.  —  I.  Cancel  the  common  factors  from  botJi  the 
dividend  and  the  divisor. 

II.  Then  divide  the  product  of  the  remaining  factors  of  the 
dividend  by  the  product  of  the  remaining  factors  of  the 
divisor,  and  the  result  will  be  the  quotient. 


EXAMPLES    FOR    PRACTICE. 
7O.      Divide: 

(a)  14  X  18  X  16  X  40  by  7  X  8  X  6  X  5  X  3. 

(b)  3  X  65  X  50  X  100  X  60  by  30  X  60  X  13  X  10. 

(c)  8X4X3X9XH  by  11X9X4X3X8. 

(d)  164x321x6x7x4by82x321x7. 

(e)  50  X  100  X  200  X  72  by  1,000  X  144  X  100. 
(/)    48  X  63  X  55  X  49  by  7  X  21  X  11  X  48. 
(g)    110  X  150  X  84  X  32  by  11  X  15  X  100  X  64. 


(a)  32. 

(b)  250. 

(c)  1. 

(d)  48. 

(e)  5. 
(/)  105. 

42. 


(h)     115  X  120X400  XI, 000  by  23X1,000X60X800.          (_(//)     5. 


ARITHMETIC. 
FHACTIOlSrS. 


DEFINITIONS. 

71.  A  fraction   is  a  part  of  a   unit.      One-half,   one- 
third,  two- fifths  are  fractions. 

72.  Two  numbers  are  required  to  express  a  fraction ;  one 
is  called  the  numerator,  and  the  other,  the  denominator. 

73.  The  numerator  is  placed  above  the  denominator, 
with  a  line  between  them,   as  -f.     Here,   3  is  the  denom- 
inator, and  shows  into  how  many  • equal  parts  the  unit,  or 
one,  is  divided.     The  numerator,    2,    shows  how  many  of 
these  equal  parts  are  taken  or  considered.     The  denomi- 
nator also  indicates  the  names  of  the  parts. 

\  is  read  one-half. 

j-  is  read  three-fourths. 

-f-  is  read  three-eighths, 
-j^-  is  read  five-sixteenths, 
-f-f-  is  read  twenty-nine  forty-sevenths. 

74.  In  the  expression  "f  of  an  apple,"  the  denominator, 
4,  shows  that  the  apple  is  to  be  (or  has  been)  cut  into  4  equal 
parts,  and  the  numerator,  3,  shows  that  tJiree  of  these  parts, 
or  fourths,  are  taken  or  considered. 

If  each  of  the  parts,  or  fourths,  of  the  apple  were  cut 
into  tivo  equal  pieces,  there  would  then  be  twice  as  many 
pieces  as  before,  or  4x2  =  8  pieces  in  all,  one  of  these 
pieces  would  be  called  one-eighth,  and  would  be  expressed 
in  figures  as  \.  Three  of  these  pieces  would  be  called 
three-eighths,  and  written  f.  The  words  three-fourths, 
three-eighths,  five-sixteenths,  etc.  are  abbreviations  of  three 
one-fourths,  three  one-eighths,  five  one-sixteenths,  etc.  It 
is  evident  that  the  larger  the  denominator,  the  greater  is  the 
number  of  parts  into  which  anything  is  divided;  conse- 
quently, the  parts  themselves  are  smaller,  and  the  value  of 
the  fraction  is  less  for  the  same  number  of  parts  taken.  In 
other  words,  •£,  for  example,  is  smaller  than  J,  because  if  an 
object  be  divided  into  9  parts,  the  parts  are  smaller  than  if 


§  1  ARITHMETIC.  23 

the  same  object  had  been  divided  into  8  parts ;  and,  since  ^ 
is  smaller  than  -|-,  it  is  clear  that  7  one-ninths  is  a  smaller 
amount  than  7  one-eighths.  Hence,  also,  f  is  less  than  {. 

75.  The  value  of  a  fraction  is  the  numerator  divided 
by  the  denominator,  as  f  —  2,  f  =  3. 

76.  The  line  between  the  numerator  and  the  denomi- 
nator means  divided  by,  or  -=-. 

\  is  equivalent  to  3  -f-  4. 
I  is  equivalent  to  5  -s-  8. 

77.  The  numerator  and  the  denominator  of  a  fraction 
are  called  the  terms  of  a  fraction. 

78.  The  value  of  a  fraction  whose  numerator  and  denom- 
inator are  equal  is  1. 

f,  or  four-fourths  =  1. 
f,  or  eight-eighths  =  1. 
If,  or  sixty-four  sixty-fourths  =  1. 

» 

79.  A  proper  fraction  is  a  fraction  whose  numerator 
is  less  than  its  denominator.     Its  value  is  less  than  1,   as 

f ,  i,  A- 

80.  An  improper  fraction  is  a  fraction  whose  numer- 
ator equals  or  is  greater  than  the  denominator.     Its  value  is 
1  or  more  than  1,  as  f ,  |,  f  f . 

81.  A  mixed  number  is  a  whole  number  and  a  fraction 
united.     4f  is  a  mixed  number,  and  is  equivalent  to  4  -\-  f . 
It  is  read  four  and  two-thirds. 


REDUCTION  OF  FRACTIONS. 

82.  Reduction  of  fractions  is  the  process  of  changing- 
their  form'  without  changing  their  value. 

83.  A  fraction  is  reduced  to  higher  terms  by  multiplying 
both  terms  of  the  fraction  by  the  same  number.     Thus,  J  is 
reduced  to  |  by  multiplying  both  terms  by  2. 

3X2  __  6 
4X2    .    8* 

1-3 


24:  ARITHMETIC.  §  1 

The  value  is  not  changed,  since  f  =  |.  For,  suppose 
that  an  object,  say  an  apple,  is  divided  into  8  equal  parts. 
If  these'  parts  be  arranged  into  4  piles,  each  containing  2 
parts,  it  is  evident  each  pile  will  be  composed  of  the  same 
amount  of  the  entire  apple  as  would  have  been  the  case  had 
the  apple  been  originally  cut  into  4  equal  parts.  Now,  if 
one  of  these  piles  (containing  2  parts)  be  removed,  there  will 
be  3  piles  left,  each  containing  2  equal  parts,  or  G  equal  parts 
in  all,  i.  e. ,  six-eighths.  But,  since  one  pile,  or  one-quarter, 
was  removed,  there  are  three-quarters  left.  Hence,  f-  =  f . 
The  same  course  of  reasoning  may  be  applied  to  any  similar 
case.  Therefore,  multiplying  both  terms  of  a  fraction  by 
the  same  number  does  not  alter  its  value. 

84.  To  reduce  a  fraction  to  an  equal  fraction  hav- 
ing a  given  denominator : 

EXAMPLE. — Reduce  £•  to  an  equal  fraction  having  96  for  a  denominator. 

SOLUTION. — Both  the  numerator  and  the  denominator  must  be  mul- 
tiplied by  the  same  number  in  order  not* to  change  the  value  of 
the  fraction.  The  denominator  must  be  multiplied  by  some  number 
which  will,  in  this  case,  make  the  product  96 ;  this  number  is  evidently 

7  v  12        84 

96  -f-  8  =  12,  since  8  X  12  =  96.     Hence,  i* '<*=*'£.     Ans. 

o  X  J-*        "V 

85.  Rule. — Divide  the  given  denominator  by  the  denom- 
inator of  the  given  fraction,  and  multiply  both  terms  of  the 
fraction  by  the  result. 

EXAMPLE.— Reduce  f  to  lOOths. 

SOLUTION.—    100-*- 4  =  25;  hence,  |  X  ^  =  ^.     Ans. 

86.  A  fraction  is  reduced  to  loiver  terms  by  dividing 
both  terms  by  the  same  number.     Thus,  T^  is  reduced  to  f 
by  dividing  both  terms  by  2. 

_8_^-2    _  4 
10-h2  Z"  5' 

That  -f$  =  \  is  readily  seen  from  the  explanation  given 
in  Art.  83 ;  for,  multiplying  both  terms  of  the  fraction  f  by 

2>  i  x  I  =  TT>>  and>  if  I  =-yV>  A  must  equal  f  Hence, 
dividing  both  terms  of  a  fraction  by  the  same  number  does 
not  alter  its  value. 


§  1  ARITHMETIC.  25 

87.  A  fraction  is  reduced  to  its  lowest  terms  when  its 
numerator  and  denominator  cannot  both  be  divided  by  the 
same  number  without  a  remainder ;  for  example,  f ,  f ,  \\,  T8T. 


EXAMPLES  FOB  PRACTICE 

88.      Reduce  the  following: 
(a)     T\  to  128ths. 
(£)     ^  to  its  lowest  terms. 

to  its  lowest  terms.  Ans. 


(d)  f  to  49ths. 

(e)  jf  to  10,000ths. 


(') 


(') 


If- 


89.  To  reduce  a  whole  number  or  a  mixed  number 
to  an  improper  fraction  : 

EXAMPLE. — How  many  fourths  in  5  ? 

SOLUTION.— Since  there  are  4  fourths  in  1  (£  =  1),  in  5  there  will  be 
5  X  4  fourths,  or  20  fourths ;  i.  e. ,  5  X  f  =  \°-     Ans. 
EXAMPLE. — Reduce  8f  to  an  improper  fraction. 
SOLUTION.—    8  X  |  =  -7--     \2-  +  l  =  -3¥5.     Ans. 

90.  Rule. — Multiply  the  wJiole  number  by  the  denomina- 
tor of  the  fraction,  add  the  numerator  to  the  product  and 
place  the  denominator  under  the  result.     If  it  is  desired  to 
reduce  a  ivhole  number  to  a  fraction,  multiply  the  whole  number 
by  the  denominator  of  the  given  fraction,  and  write  the  result 
over  the  denominator. 

EXAMPLES  FOR  PRACTICE. 

91.  Reduce  to  improper  fractions : 

(a) 

(b)      5£. 

w 


(d)  37f.  Ans<  * 

(e)  60|. 

(/)    Reduce  7  to  a  fraction  whose  denominator  is  16.  I  (/) 


92.  To  reduce  an  improper  fraction  to  a  \vhole  or 
a  mixed  number: 

EXAMPLE. — Reduce  ^  to  a  mixed  number. 

SOLUTION. —  4  is  contained  in  21,  5  times  and  1  remaining  (see  Art. 
75) ;  as  this  is  also  divided  by  4,  its  value  is  £.  Therefore,  5  +  J,  or  5£, 
is  the  number.  Ans. 


26  ARITHMETIC.  §  1 

93.  Rule. — Divide  the  numerator  by  the  denominator, 
the  quotient  will  be  the  whole  number;   the  remainder,   if 
there  be  any,  will  be  the  numerator  of  the  fractional  part  of 
which  the  denominator  is  the  same  as  the  denominator  of  the 
improper  fraction. 

EXAMPLES  FOR  PRACTICE. 

94.  Reduce  to  whole  or  mixed  numbers: 


(«) 


JLtJL- 


(')     It- 
(/)   W- 


(a)  24£. 

(b)  61f 


Ans"  ^   (d)     49f. 
(/)    5. 


95.  A  common  denominator  of  two  or  more  fractions 
is  a  number  which  will  contain  (i.  e. ,  which  may  be  divided 
by)  the  denominator  of  each  of  the  given  fractions  without 
a  remainder.     The   least   common   denominator  is  the 

least  number  that  will  contain  each   denominator  of  the 
given  fractions  without  a  remainder. 

96.  To  find  the  least  common  denominator: 

EXAMPLE. — Find  the  least  common  denominator  of  £,  ^,  £,  and  T^. 
SOLUTION. — We  first  place  the  denominators  in  a  row,  separated  by 
commas.  2  )  4,     3,     9,  16 

2  )  2,     3,     9,     8 

3  7T     ^     9^    4 

3  )  1,     1,     3,     4 

4  )  1,     1,     1,     4 

1,     1,     1,     1 
2X2X3X3X4  =  144,  the  least  common  denominator.     Ans. 

EXPLANATION. — Divide  each  of  them  by  some  prime  num- 
ber which  will  divide  at  least  two  of  them  without  a 
remainder  (if  possible),  bringing  down  those  denominators 
to  the  row  below  which  will  not  contain  the  divisor  without 
a  remainder.  Dividing  each  of  the  numbers  by  2,  the  sec- 
ond row  becomes  2,  3,  9,  8,  since  2  will  not  divide  3  and  9 
without  a  remainder.  Dividing  again  by  2,  the  result  is  1, 
3,  9,  4.  Dividing  the  third  row  by  3,  the  result,  is  1,  1, 


§  1  ARITHMETIC.  27 

3,  4.  So  continue  until  the  last  row  contains  only  1's. 
The  product  of  all  the  divisors,  or  2x2x3x3x4  =  144,  is 
the  least  common  denominator. 

97.  EXAMPLE. — Find  the  least  common  denominator  of  f,  ^,  T7¥. 
SOLUTION.—  3  )  9,  12,  18 

3  )  3,  4,     6 

2)1,  4.     2 

2  )  1,  2,     1 

1,  1,     1 

3X3X2X2  =  36.     Ans. 

98.  To  reduce  two  or  more  fractions  to  fractions 
having  a  common  denominator: 

EXAMPLE. — Reduce  f ,  f ,  and  £  to  fractions  having  a  common  denom- 
inator. 

SOLUTION. — The  common  denominator  is  a  number  which  will  con- 
tain 3,  4,  and  2.  The  least  common  denominator  is  12,  because  it  is 
the  smallest  number  which  can  be  divided  by  3,  4,  and  2  without  a 

remainder.  2  _     s       3—9       i  _     6 

•3  —  ~i~s>     t  —  T¥>     ¥  —  Ti~- 

Reducing  f  (see  Art.   84),  3  is  contained  in  12,  4  times.     By  multi- 
plying both  numerator  and  denominator  of  f  by  4,  we  find 
2X4        8 
3X4  =  12-     Intl 

99.  Rnle. — Divide    the    common    denominator    by    the 
denominator  of  the  given  fraction,  and  multiply  botJi  terms 
of  the  fraction  by  the  quotient. 


EXAMPLES   FOR  PRACTICE. 

1OO.      Reduce  to  fractions  having  a  common  denominator; 
(«)      f .  f .  f 


Ans 


(')  -A.  A.  A- 
(/)   TVU.fi 


(')     if.  A.  if • 


ADDITION    OF    FRACTIONS. 

1O1.  Fractions  cannot  be  added  unless  they  have  a  com- 
mon denominator.  We  cannot  add  f  to  \  as  they  now  stand, 
since  the  denominators  represent  parts  of  different  sizes. 
Fourths  cannot  be  added  to  eighths. 


28  ARITHMETIC.  §  1 

Suppose  we  .divide  an  apple  into  4  equal  parts,  and  then 
divide  2  of  these  parts  into  2  equal  parts.  It  is  evident 
that  we  shall  have  2  one-fourths  and  4  one-eighths.  Now, 
if  we  add  these  parts,  the  result  is  2  +  4  =  6  something. 
But  what  is  this  something  ?  It  is  not  fourths,  for  6 
fourths  are  1|-,  and  we  had  only  1  apple  fo  begin  with; 
neither  is  it  eighths,  for  6  eighths  are  f  ,  which  is  less  than 
1  apple.  By  reducing  the  quarters  to  eighths,  we  have 
f  =  |-,  and  adding  the  other  4  eighths,  4  +  4  =  8  eighths. 
This  result  is  correct,  since  -J  =  1.  Or  we  can,  in  this  case, 
reduce  the  eighths  to  quarters.  Thus,  |-  =  \  ;  whence, 
adding,  2  +  2  =  4  quarters,  a  correct  result,  since  \  •=.  1. 

Before  adding,  fractions  should  be  reduced  to  a  common 
denominator,  preferably  the  least  common  denominator. 

1O2.      EXAMPLE.—  Find  the  sum  of  £,  f,  and  f. 

SOLUTION.  —  The  least  common  denominator,  or  the  least  number 
which  will  contain  all  the  denominators,  is  8. 

i  =  t,     1  =  1.     and    |  =  f  . 

EXPLANATION.  —  As  the  denominator  tells  or  indicates  the 
names  of  the  parts,  the  numerators  only  are  added,  to  obtain 
the  total  number  of  parts  indicated  by  the  denominator. 
Thus,  4  one-eighths  plus  6  one-eighths  plus  5  one-eighths  = 


1O3.      EXAMPLE.—  What  is  the  sum  of  12|,  14f,  and  7T5¥? 
SOLUTION.  —  The  least  common  denominator  in  this  case  is  16. 

12*  =  12H 
14f  =  14ft 

7*  =  _*&_ 

sum  '  33  +  fl  =  33  +  1ft  =  34ft.    Ans. 

The  sum  of  the  fractions  =  f  £  or  1ft,  which  added  to  the  sum  of  the 
whole  numbers  =  34ft. 

EXAMPLE.—  What  is  the  sum  of  17,  13T\,  &,  and  8J? 
SOLUTION.  —  The  least  common   denominator  is    32.     13^  = 
Si  =  8£.  17 

18* 


_ 

sum    33||.     Ans. 


§  1  ARITHMETIC.  29 

1O4.  Rule. — I.  Reduce  the  given  fractions  to  frac- 
tions having  tJie  least  common  denominator,  and  write  the 
sum  of  the  numerators  over  the  common  denominator. 

II.  When  there  are  mixed  numbers  and  whole  numbers, 
add  tJie  fractions  first,  and  if  their  sum  is  an  improper 
fraction,  reduce  it  to  a  mixed  number  and  add  the  whole 
number  with  the  other  whole  numbers. 


EXAMPLES  FOR  PRACTICE. 

1O5.      Find  the  sum  of: 

(«)     f  A,f- 
(*)      *,  T^  If- 

(0  *,.!,  fV 

(<0  i  ii  if-  A 

(')  if.sV  If. 

(/)  ft,  iiif 

(£•)  A.  A.  if 

<*)  f.if.  f 


(b}  IA- 

(0  ii3^- 

W  iff- 

(/)  if- 

(/o  i!" 


SUBTRACTION    OF    FRACTIONS. 

1O6.  Fractions  cannot  be  subtracted  ivithout  first  re- 
ducing them  to  a  common  denominator.  This  can  be 
shown  in  the  same  manner  as  in  the  case  of  addition  of 
fractions. 

EXAMPLE. — Subtract  f  from  ||. 
SOLUTION. — The  common  denominator  is  16. 

13-6 


J(f 


Ans. 


107.  EXAMPLE. — From  7  take  |. 

SOLUTION.—    1  =  f ;  therefore,  since  7  =  6  +  1,  7  =  6  +  f  =  6f ,  or 
—  |  =  6f.     Ans. 

108.  EXAMPLE.— What  is  the  difference  between  17^  and  9||  ? 
SOLUTION. — The  common  denominator  of  the  fractions  is  32.     17^ 


minuend 

stibtrahend 

difference 


Ans. 


30  ARITHMETIC.  §  1 

1O9.      EXAMPLE.—  From  9£  take  4TV 

SOLUTION.  —  The  common  denominator  of  the  fractions  is  16.     9^ 

9iV  minuend    9T\  or  8f  £ 

subtrahend    4^      4T7^ 
dijference  ~4£f     ~4£f.     Ans. 

EXPLANATION.  —  As  the  fraction  in  the  subtrahend  is 
greater  than  the  fraction  in  the  minuend,  it  cannot  be  sub- 
tracted; therefore,  borrow  1,  or  if,  from  the  9  in  the 
minuend  and  add  it  to  the  T\;  -j^-hyf  —  TIP  yV  fr°m 
|f  =  if.  Since  1  was  borrowed  from  9,  8  remains;  4  from 
8  =  4;  4  +  lf  =  4if. 


110.  EXAMPLE.—  From  9  take  8T\. 

SOLUTION.  —  minuend    9     or 

subtrahend    8T\ 
difference      f|        £f.     Ans. 

EXPLANATION.  —  As  there  is  no  fraction  in  the  minuend 
from  which  to  take  the  fraction  in  the  subtrahend,  borrow 
1,  or  if,  from  9.  T3g-  from  if  —  if.  Since  1  was  borrowed 
from  9,  only  8  is  left.  8  from  8  =  0. 

111.  Rule.  —  I.     Reduce    the    fractions     to    fractions 
having1  a  common   denominator.       Subtract   one    numerator 
from  the  other  and  place  the  remainder  over  the  common 
denominator. 

II.  When  there  are  mixed  numbers,  subtract  the  fractions 
and  whole  numbers  separately,  and  place  the  remainders  side 
by  side. 

III.  When   the  fraction   in   the  subtrahend  is   greater 
than  the  fraction  in  the  minuend,  borrow  1  from  the  ivhole 
number   in   the    minuend  and  add    it   to   the  fraction    in 
the  minuend,  from  which  subtract  the  fraction  in  the  sub- 
trahend. 

IV.  When  the  minuend  is  a  whole  number,  borrow  1; 
reduce  it  to  a  fraction  whose  denominator  is  the  same  as  the' 
denominator  of  the  fraction  in  the  subtralicnd,  and  place  it 
over  that  fraction  for  subtraction. 


81 


ARITHMETIC. 


31 


EXAMPLJE8  FOll  PRACTICE. 
5.      Subtract: 
(a)     ££  from  £f 
(&)      ^  from  ^f 
(£•)      ¥47  from  T5^. 
(rf)     ^  from  ff 
(*)       if  from  f  f 
(/)    13±  from  30£. 
(£•)    12£  from  27. 
(/*)     5i  from  30. 


(d) 
(*) 

to 

(d?> 

to 


14*. 


MUIjTIPilCATION    OF    FRACTIONS. 

113.  /;/  multiplication  of  fractions  it  is  not  necessary  to 
reduce  tJie  fractions  to  fractions  having  a  common  denominator. 

114.  Multiplying  the  numerator  or  dividing  the  denom- 
inator multiplies  the  fraction, 

EXAMPLE.  —  Multiply  f  by  4. 

SOLUTION.—  x4  =     X4  =  J     =  3.    Ans. 


Or,     f  X4  =     ^_4  ==  f  =  3. 


Ans. 


The  word  "of,"  when  placed  between  two  fractions,  or 
between  a  fraction  and  a  whole  number,  means  the  same  as 
X ,  or  times.  Thus, 

|  of  4  =  |X4  =  3. 

EXAMPLE. — Multiply  f  by  2. 

SOLUTION. —  2X1  =  3         =  I  =  f  •     Ans. 

o 
o 

Or,     2  X I  =  g  ^  o  =  £•     Ans- 

115.      EXAMPLE. — What  is  the  product  of  T\  and  |  ? 
SOLUTION. —       A  X  „ 
I 


4  V  7 

=  i  X  i  =  A\  =  A.     Ans. 


Or,  by  cancelation, 


4X8 


=.  A-    Ans- 


116.      EXAMPLE.— What  is  f  of  f  of  ||  ? 


SOLUTION. — 


t  x  3  x  n  _    3. 

8X^X3?  "~  8X2 


=  A.     Ans. 


ARITHMETIC. 


EXAMPLE.— What  is  the  product  of  9f  and  5f  ? 
SOLUTION.—  9f  =  ^ ;  5|  =  \5-. 

¥X¥  =  f  *f  =  HF  =  54M-   Ans' 

118.  EXAMPLE.  —Multiply  15|  by  3. 
SOLUTION. —  15£  15| 

_3     or     3 

47f  45  +  \L  =  45  +  2f  =  47f .     Ans. 

119.  Rule. — I.     Divide  the  product  of  the  numerators 
by  the  product  of  the  denominators.     All  factors  common  to 
the  numerators  and  denominators  should*  first  be  cast  out  by 
cancelation. 

II.  To  multiply  one  mixed  number  by  another,  reduce  them 
both  to  improper  fractions. 

III.  To  multiply  a  mixed  number  by  a  whole  number,  first 
multiply  the  fractional  part  by  the  multiplier,  and  if  the  prod- 
uct is  an  improper  fraction,  reduce  it  to  a  mixed  number 
and  add  the  whole-number  part  to  the  product  of  the  mul- 
tiplier and  the  whole  number. 


EXAMPLES  FOR  PRACTICE. 
1 2O.     Find  the  product  of : 

(«)     7  X  TV 
(b)      14  X  TV 

if. 
4. 


Ans. 


(/)     17HX7. 
(jsr)    ill  X  32. 


(/)    125. 


DIVISION    OF    FRACTIONS. 

In  division  of  fractions  it  is  not  necessary  to  reduce 
the  fractions  to  fractions  having  a  common  denominator. 

122.     Dividing  the  numerator  or  multiplying  the  denom- 
inator divides  the  fraction. 

EXAMPLE. — Divide  f  by  3. 

SOLUTION. — When  dividing  the  numerator,  we  have 

^8  =         3  =      =         Ans. 


1  ARITHMETIC.  33 

When  multiplying  the  denominator,  we  have 

A 

iK3  =  gx3  =  -&  =  \.    Ans. 
EXAMPLE.  —  Divide  T3^  by  2. 

o 

SOLUTION.—  &  -4-  2  =  ^  ^  g  =  ^V     Ans. 

EXAMPLE.  —  Divide  ^f  by  7. 

14  —  7 
SOLUTION.—  ~^7  =  =        =  Ans- 


123.  To  invert  a  fraction  is  to  /^r^  zV  upside  down  ; 
that  is,  make  the  numerator  and  denominator  change  places. 

Invert  f  and  it  becomes  f  . 

124.  EXAMPLE.—  Divide  T9W  by  TV 

SOLUTION.  —  1.  The  fraction  T3¥  is  contained  in  T9^,  3  times,  for  the 
denominators  are  the  same,  and  one  numerator  is  contained  in  the 
other  3  times.  2.  If  we  now  invert  the  divisor,  T3^,  and  multiply,  the 
solution  is 

9       16_!xtf_8      Ang 
16X  3    -  W><$ 
This  brings  the  same  quotient  as  in  the  first  case. 

125.  EXAMPLE.  —  Divide  f  by  £. 

SOLUTION.  —  We  cannot  divide  f  by  £,  as  in  the  first  case  above,  for 
the  denominators  are  not  the  same  ;  therefore,  we  must  solve  as  in  the 
second  case. 

l^f=fXf  =|^=|orli.    Ans. 
2 

126.  EXAMPLE.—  Divide  5  by  £$. 
SOLUTION.  —    -^  inverted  becomes  ^§. 


127.  EXAMPLE.  —  How  many  times  is  3f  contained  in 
SOLUTION.—  3f  =  ^;  7T7¥  =  Vir- 

•2^-  inverted  equals  T^. 
119       4        119  X^       119 
16  X15~^X15~  60  = 
4 

128.  Rule.  —  Invert  the  divisor  and  proceed  as  in  mul- 
tiplication. 


34  ARITHMETIC.  §  1 

129.  We  have  learned  that  a  line  placed  between  two 
numbers  indicates  that  the  number  above  the  line  is  to  be 
divided  by  the  number  below  it.     Thus,  ±£-  shows  that  18 
is  to  be  divided  by  3.     This  is  also  true  if  a  fraction  or  a 
fractional  expression  be  placed  above  or  below  a  line. 

9  3x7 

-  means  that  9  is  to  be  divided  by  £ ;  means  that 

•§•  8 -T-4 

16 

3  X  7  is  to  be  divided  by  the  value  of     ?~   . 

2  is  the  same  as  i  -=-  £. 

8 

It  will  be  noticed  that  there  is  a  heavy  line  between  the  9 
and  the  £.  This  is  necessary,  since  otherwise  there  would  be 
nothing  to  show  as  to  whether  9  was  to  be  divided  by  £,  or 
f  was  to  be  divided  by  8.  Whenever  a  heavy  line  is  used, 
as  shown  here,  it  indicates  that  all  above  the  line  is  to  be 
divided  by  all  below  it. 

EXAMPLES   FOR  PRACTICE. 

130.  Divide: 


(a)  15by6f. 

(*)  30byf. 

(c)  172  by  f  . 

(</)  lfbylTV 

(e)  103.  by  14f  . 

(/)  Wbyif*- 
(*) 

(k) 


(a)  2£. 

(b)  40. 

(c)  215. 

W  Mf 

00  Hi 


131.  Whenever  an  expression  like  one  of  the  three 
following  ones  is  obtained,  it  may  always  be  simplified  by 
transposing  the  denominator  from  above  to  below  the  line, 
or  from  below  to  above,  as  the  case  may  be,  taking  care, 
however,  to  indicate  that  the  denominator  when  so  trans- 
ferred is  a  multiplier. 
SL  3 

—  A  —  IT  '    f°r>    regarding    the    fraction 


9X4 
above  the  heavy  line  as  the  numerator  of  a  fraction  whose 

denominator  is  9,  \  — ,  as  before. 

J  X  4         J  X  4 


§  1  ARITHMETIC.  35 

9        9x4 

2.     -  =  —5  —  =  12.    The  proof  is  the  same  as  in  the  first 

f  o 


case. 


t5x4 
=  Q  —  o  =  IT-     For,  regarding  %  as  the  numerator 

^      o  x  y 

of  a  fraction  whose  denominator  is  £  ,  |        =  -  —  -  ;    and 

if  x  y      o  x  y 

5    X4        5X4 


4 
This  principle  may  be  used  to  great  advantage  in  cases 

.     Reducing    the    mixed    numbers  to 


r      ..  .  T,T 

fractions,  the  expression   becomes  4       —    1231  -  .      Now 

40  X  Y  X  -g- 

transf  erring  the  denominators  of  the  fractions  and  canceling, 

3 

n   s    0      3 

1x310x27x72x2x6 
40X9X31X4X12 


Greater  exactness  in  results  can  usually  be  obtained  by 
using  this  principle  than  by  reducing  the  fractions  to  deci- 
mals. The  principle,  however,  should  not  be  employed  if 
a  sign  of  addition  or  subtraction  occurs  either  above  or 
below  the  dividing  line. 


DECIMALS. 

132.  Decimals  are  tenth  fractions;  that  is,  the  parts  of 
a  unit  are  expressed  on  the  scale  of  ten,  as  tenths^  hun- 
dredths,  thousandths,  etc. 

133.  The  denominator  which  is  always  ten  or  a  multiple 
of  ten,  as  10,  100,  1,000,  etc.,  is  not  expressed,  as  it  would 


36  ARITHMETIC.  §  1 

be  in  common  fractions,  by  writing-  it  tinder  the  numerator 
with  a  line  between  them,  as  y3^,  Tf  ^  -j-^,  but  is  expressed 
by  placing  a  period  ( .),  which  is  called  a  decimal  point,  to 
the  left  of  the  figures  of  the  numerator •,  so  as  to  indicate 
that  the  number  on  the  right  is  the  numerator  of  a  fraction 
whose  denominator  is  10,  100,  1,000,  etc. 

134.  The  reading  of  a  decimal  number  depends  upon 
the  number  of  decimal  places  in  it,  or  the  number  of  figures 
to  the  right  of  the  decimal  point. 

One  decimal  place  expresses  tenths. 

Two  decimal  places  express  hundredths. 

Three  decimal  places  express  thousandths. 

Four  decimal  places  express  ten-thousandths. 

Five  decimal  places  express  hundred-thousandths. 

Six  decimal  places  express  millionth*. 

Thus: 

.3  =      A-  =  3  tenths. 

.03  =      rihr  =  3  hundredths. 

.003  =     -j-^pj-  =  3  thousandths. 

.0003  =    TTnnnr  =  3  ten-thousandths. 

.00003  =  -nnnnnr  =  3  hundred-thousandths. 

.000003  —  y^nmnnr  =  3  million ths. 

We  see  in  the  above  that  the  number  of  decimal,  places  in 
a  decimal  equals  the  number  of  ciphers  to  the  right  of  the 
figure  1  in  the  denominator  of  its  equivalent  fraction.  This 
fact  kept  in  mind  will  be  of  much  assistance  in  reading  and 
writing  decimals. 

Whatever  may  be  written  to  the  left  of  a  decimal 
point  is  a  whole  number.  The  decimal  point  merely  sep- 
arates the  fraction  on  the  right  from  the  whole  number  on 
the  left. 

When  a  whole  number  and  decimal  are  written  together, 
the  expression  is  a  mixed  number.  Thus,  8. 12  and  17.25  are 
mixed  numbers. 


g  1  ARITHMETIC.  37 

The  relation  of  decimals  and  whole  numbers  to  each  other 
is  clearly  shown  by  the  following  table : 


The  figures  to  the  left  of  the  decimal  point  represent 
whole  numbers;  those  to  the  right  are  decimals. 

In  both  the  decimals  and  whole  numbers,  the  units  place 
is  made  the  starting  point  of  notation  and  numeration. 
Both  whole  numbers  and  decimals  decrease  on  the  scale  of 
ten  to  the  right,  and  both  increase  on  the  scale  of  ten  to  the 
left.  The  first  figure  to  the  left  of  units  is  tens,  and  the 
first  figure  to  the  rigJit  of  units  is  tenths.  The  second  figure 
to  the  left  of  units  is  hundreds,  and  the  second  figure  to  the 
'  rigJit  is  hundredths.  The  third  figure  to  the  left  is  thousands ^ 
and  the  third  to  the  right  is  thousandths^  and  so  on ;  the 
iv hole  numbers  on  the  left  and  the  decimals  on  the  right. 
The  figures  equally  distant  from  units  place  correspond  in 
name,  the  decimals  having  the  ending  ths,  to  distinguish 
them  from  whole  numbers.  The  following  is  the  numeration 
of  the  number  in  the  above  table :  nine  hundred  eighty-seven 
million,  six  hundred  fifty-four  thousand,  three  hundred 
twenty-one  and  twenty-three  million,  four  hundred  fifty-six 
thousand,  seven  hundred  eighty-nine  hundred-millionths. 

The  decimals  increase  to  the  left,  on  the  scale  of  ten,  the 
same  as  whole  numbers;  for,  if  you  begin  at  the  4  in 
thousandths  place  in  the  above  table,  the  next  figure  to  the 
left  is  hundredths,  which  is  ten  times  as  great,  and  the  next 
tenths,  or  ten  times  the  hundredths,  and  so  on  through  both 
decimals  and  whole  numbers. 


38  ARITHMETIC.  §  1 

135.  Annexing^  or  taking  aivay,  a  cipher  at  the  right  of 
a  decimal,  does  not  affect  its  value. 

.5  is  ^;  .50  is  ^V  but  T5o  =  f^r  ;  therefore,  .5  =.50. 

136.  Inserting  a  cipher  between  a  decimal  and  the  decimal 
point,  divides  the  decimal  by  10. 

•5  =  A;  A-*-  10  =  ^=.05. 

137.  Taking  away  a  cipher  from  the  left  of  a  decimal, 
multiplies  the  decimal  by  10. 

lO  -         =.5. 


138.  In  some  cases  it  is  convenient  to  express  a  mixed 
decimal  fraction  in  the  form  of  a  common  (improper)  frac- 
tion. To  do  so  it  is  only  necessary  to  write  the  entire  num- 
ber, omitting1  the  decimal  point,  as  the  numerator  of  the 
fraction,  and  the  denominator  of  the  decimal  part  as  the 
denominator  of  the  fraction.  Thus,  127.483  =  ^-f^f1;  for, 
127.483  = 


ADDITION    OF    DECIMALS. 

139.  Addition  of  decimals  is  similar  in  all  respects  to 
addition  of  whole  numbers — units  are  placed  under  units, 
tens  under  tens,  etc. ;  this,  of  course,  brings  the  decimal 
points  in  line,  directly  under  one  another.  Hence,  in  pla- 
cing the  numbers  to  be  added,  it  is  only  necessary  to  take 
care  that  the  decimal  points  are  in  line.  In  adding  whole 
numbers,  the  right-hand  figures  are  always  in  line;  but 
in  adding  decimals,  the  right-hand  figures  will  not  be  in 
line  unless  each  decimal  contains  the  same  number  of 
figures. 

wltole  numbers  decimals  mixed  numbers 

342  .342  342.032 

4234  .4234  4234.5 

26  .26  26.6782 

3_  .03  3.06 

sum    4605    Ans.  sum    1.0  5  5  4  Ans.      sum    4  6  0  6.2  7  0  2    Ans. 


ARITHMETIC. 


39 


14O.      EXAMPLE.—  What  is  the  sum  of  242,  .36,  118.725,  1.005,  6, 
and  100.1? 


SOLUTION.  — 


242. 

.36 
1  1  8.7  2  5 

1.005 

6. 
100.1 


sum 


468.190    Ans. 

141.  Rule.  —  Place  the  numbers  to  be  added  so  that  the 
decimal  points  will  be  directly  under  each  other.  Add  as 
in  wJiole  numbers,  and  place  the  decimal  point  in  the  sum 
directly  under  the  decimal  points  above. 


EXAMPLES  FOR  PRACTICE. 

Find  the  sum  of: 

(a)  .2143,  .105,  2.3042,  and  1.1417. 

(£)  783.5,  21.473,  .2101,  and  .7816. 

(c)  21.781,  138.72,  41.8738,  .72,  and  1.413. 

(d)  .3724,  104.15,  21.417,  and  100.042. 

(*)  200.172,  14.105,  12.1465,  .705,  and  7.2. 

(/)  1,427.16,  .244,  .32,  .032,  and  10.0041. 

(£•)  2,473.1,  41.65,  .7243,  104.067,  and  21.073. 

(/i)  4,107.2,  .00375,  21.716,  410.072,  and  .0345. 


» 
w 

M 

(<*) 
M 

00 
te) 
.(*) 

3.7652. 
805.9647. 
204.5078. 
225.9814. 
234.3285. 
1,437.7601. 
2,640.6143. 
4,539.02625. 

SUBTRACTION    OF    DECIMALS. 

143.  As  in  subtraction  of  whole  numbers,   units  are 
placed   under  units,   tens   under   tens,   etc.,   bringing-  the 
decimal  points-  under  each  other,  as  in  addition  of  decimals. 

EXAMPLE.— Subtract  .132  from  .3063. 
SOLUTION. —  minuend    .3063 

subtrahend    .132 
difference    .1743    Ans.     . 

144.  EXAMPLE.— What  is  the  difference  between  7.895  and  .725  ? 
SOLUTION. —  minuend    7.8  9  5 

subtrahend      .725 
difference    7.1  70  or  7.1  7.     Ans. 


40  ARITHMETIC.  §  1 

145.  EXAMPLE.— Subtract  .625  from  11. 
SOLUTION. —  minuend    1  1.0  0  0 

subtrahend         .625 
difference    1  0.3  7  5    Ans. 

146.  Rule. — Place  the  subtrahend  under  the  minuend, 
so  that  the  decimal  points  will  be  directly  under  each  other. 
Subtract  as  in  whole  numbers,  and  place  the  decimal  point  in 
the  remainder  directly  under  the  decimal  points  above. 

When  the  figures  in  the  decimal  part  of  the  subtrahend 
extend  beyond  those  in  the  minuend,  place  cipliers  in  the  min- 
uend above  them  and  subtract  as  before. 


EXAMPLES  FOR  PRACTICE. 
147.      From: 


(a)  407.385  take  235.0004. 

(£)  22. 718  take  1.7042. 

(c)  1,368. 17  take  13. 6817. 

(d)  70.00017  take  7.000017. 
(<?)  630.630  take  .6304. 
(/)  421.73  take  217.162. 
(g)  1.000014  take  .00001. 
(h)  .783652  take  .542314. 


Ans. 


(a)  172.3846. 

(b)  21.0138. 

(c)  1,354.4883. 

(d)  63.000153. 
(*)  629.9996. 
(/)  204.568. 
(£•)  1.000004. 
(//)  .241338. 


MULTIPLICATION   OF   DECIMALS. 

148.  In  multiplication  of  decimals  we  do  not  place  the 
decimal  points  directly  under  each  other  as  in  addition  and 
subtraction.  We  pay  no  attention  for  the  time  being  to  the 
decimal  points.  Place  the  multiplier  under  the  multiplicand, 
so  that  the  right -hand  figure  of  the  one  is  under  the  right- 
hand  figure  of  the  other,  and  proceed  exactly  as  in  multipli- 
cation of  whole  numbers.  After  multiplying,  count  the 
number  of  decimal  places  in  both  multiplicand  and  multiplier, 
and  point  off  the  same  number  in  the  product. 

.    EXAMPLE.— Multiply  .825  by  13. 

SOLUTION. —      multiplicand         .825 
multiplier  1  3 

2475 
825 
product    10.725    Ans. 


§  1  ARITHMETIC.  41 

In  this  example  there  are  3  decimal  places  in  the  multi- 
plicand and  none  in  the  multiplier;  therefore,  3  decimal 
places  are  pointed  off  in  the  product. 

1 49.  EXAMPLE.— What  is  the  product  of  426  and  the  decimal . 005  ? 
SOLUTION. —        multiplicand       426 

multiplier      .005 

product    2.1  30  or  2. 13.     Ans. 

In  this  example  there  are  3  decimal  places  in  the  mul- 
tiplier and  none  in  the  multiplicand;  therefore,  3  decimal 
places  are  pointed  off  in  the  product. 

150.  It  is  not  necessary  to  multiply  by  the  ciphers  on 
the  left  of  a  decimal;  they  merely  determine  the  number 
of  decimal  places.     Ciphers  to  the  rigJit  of  a  decimal  should 
be  omitted,  as  they  only  rna^e  more  figures  to  deal  with, 
and  do  not  change  the  value. 

151.  EXAMPLE.— Multiply  1.205  by  1.15. 
SOLUTION. —     multiplicand          1.2  0  5 

multiplier  1.1  5 

6025 
1205 
1205 
product    1.3  8  5  7  5    Ans. 

In  this  example  there  are  3  decimal  places  in  the  mul- 
tiplicand and  2  in  the  multiplier ;  therefore,  3  -f-  2,  or  5, 
decimal  places  must  be  pointed  off  in  the  product. 

152.  EXAMPLE.— Multiply  .232  by  .001. 
SOLUTION. —     multiplicand  .232 

multiplier  .001 

product    .000232    Ans. 

In  this  example  we  multiply  the  multiplicand  by  the  digit 
in  the  multiplier,  which  gives  232  for  the  product ;  but  since 
there  are  3  decimal  places  each  in  the  multiplier  and  multi- 
plicand, we  must  prefix  3  ciphers  to  the  232  to  make  3  -f-  3, 
or  6,  decimal  places  in  the  product. 

153.  Rule. — Place  the  multiplier  under  the  multiplicand, 
disregarding  the  position  of  the  decimal  points.     Multiply 


42  ARITHMETIC.  §  1 

as  in  whole  numbers,  and  in  the  product  point  off  as  many 
decimal  places  as  there  are  decimal  places  in  both  multiplier 
and  multiplicand,  prefixing  ciphers  if  necessary. 


EXAMPLES  FOR  PRACTICE. 

154.      Find  the  product  of : 


(a)  .000492X4.1418. 

(b)  4,003.2X1.2. 

(c)  78.6531  X  1-03. 

(d)  .3685  X  .042. 

(e)  178,352  X- 01. 
(/)  .  00045  X- 0045. 
(g)  .714  X  .00002. 
(//)  .  00004  X- 008. 


(a)  .0020377656. 

(£)  4,803.84. 

(c)  81.012693. 

(<t)  .015477. 

(e)  1,783.52. 

(/)  .000002025. 

(£•)  .00001428. 

(//)  .00000032. 


DIVISION"    OF    DECIMALS. 

155.  In  division  of  decimals  we  pay  no  attention  to  the 
decimal  point  until  after  the  division  has  been  performed. 
The  number  of  decimal  places  in  the  dividend  must  equal  (or 
be  made  to  equal  by  annexing  ciphers]  the  number  of  decimal 
places  in  the  divisor.  Divide  exactly  as  in  whole  numbers. 
Subtract  the  number  of  decimal  places  in  tJie  divisor  from  tJie 
number  of  decimal  places  in  the  dividend,  and  point  off  as 
many  decimal  places  in  the  quotient  as  there  are  units  in  the 
remainder  thus  found. 

EXAMPLE.— Divide  .625  by  25. 

divisor  dividend  quotient 
SOLUTION.—  2  5  )  .6  2  5  (  .0  2  5    Ans. 

50 

125 

1_?JL 
rcmaindcr       0 

In  this  example  there  are  no  decimal  places  in  the  divisor, 
and  three  decimal  places  in  the  dividend ;  therefore,  there 
are  3  minus  0,  or  3,  decimal  places  in  the  quotient.  One 
cipher  has  to  be  prefixed  to  the  25  to  make  the  three  decimal 
places. 


ARITHMETIC.  43 


156.      EXAMPLE.— Divide  6.035  by  .05. 

divisor   dividend   quotient 

SOLUTION. —  .0  5  )  6.0  3  5  (  1  2  0.7    Ans. 

5 

1  0 
10 


35 
35 

remainder     0 

In  this  example  we  divide  by  5,  as  if  the  cipher  were  not 
before  it.  There  is  one  more  decimal  place  in  the  dividend 
than  in  the  divisor;  therefore,  one  decimal  place  is  pointed 
off  in  the  quotient. 

157.  EXAMPLE. —Divide  .125  by  .005. 

divisor  dividend  qtiotient 

SOLUTION.—  .005).!  25(2  5    Ans. 

10 

25 
25 

remainder    0 

In  this  example  there  are  the  same  number  of  decimal 
places  in  the  dividend  as  in  the  divisor ;  therefore,  the  quo- 
tient has  no  decimal  places,  and  is  a  whole  number. 

158.  EXAMPLE.— Divide  326  by  .25. 

divisor  dividend     quotient 

SOLUTION.—  .25)326.00(1304    Ans. 

25 

76 

75 

"Too 

100 

remainder        0 

In  this  problem  two  ciphers  were  annexed  to  the  div- 
idend, to  make  the  number  of  decimal  places  equal  to 
the  number  in  the  divisor.  The  quotient  is  a  whole 
number. 


44  ARITHMETIC.  §  1 

159.  EXAMPLE.— Divide  .0025  by  1.25. 

divisor     dividend    quotient 

SOLUTION.—  1.2  5  )  .0  0  2  5  0  ( .0  0  2    Ans. 

250 
remainder        0 

EXPLANATION. — In  this  example  we  are  to  divide  .0025  by 
1.25.  Consider  the  dividend  as  a  whole  number,  i.  e.,  as  25 
(disregarding  the  two  ciphers  at  its  left,  for  the  present) ; 
also,  consider  the  divisor  as  a  whole  number,  i.  e.,  as  125. 
It  is  clearly  evident  that  the  dividend,  25,  will  not  contain 
the  divisor,  125 ;  we  must,  therefore,  annex  one  cipher  to 
the  25,  thus  making  the  dividend  250.  125  is  contained 
twice  in  250,  so  we  place  the  figure  2  in  the  quotient.  In 
pointing  off  the  decimal  places  in  the  quotient,  it  must  be 
remembered  that  there  were  only  four  decimal  places  in  the 
dividend;  but  one  cipher  was  annexed,  thereby  making 
4-f-l,  or  5,  decimal  places.  Since  there  are  five  decimal 
places  in  the  dividend  and  two  decimal  places  in  the  divisor, 
we  must  point  off  5  —  2,  or  3,  decimal  places  in  the  quotient. 
In  order  to  point  off  three  decimal  places,  two  ciphers  must 
be  prefixed  to  the  figure  2,  thereby  making  .002  the  quo- 
tient. It  is  not  necessary  to  consider  the  ciphers  at  the  left 
of  a  decimal  when  dividing,  except  when  determining  the 
position  of  the  decimal  point  in  the  quotient. 

160.  Rule. — I.    Place  the  divisor  to  the  left  of  the  divi- 
dend, arid  proceed  as  in  division  of  ivhole  numbers;  in  the 
quotient,  point  off  as  many  decimal  places  as  the  number  of 
decimal  places  in  the  dividend  exceed  those  in  the  divisor,  pre- 
fixing ciphers  to  the  quotient,  if  necessary. 

II.  If  in  dividing  one  number  by  another  there  be  a 
remainder,  the  remainder  can  be  placed  over  the  divisor, 
as  a  fractional  part  of  the  quotient,  but  it  is  generally 
better  to  annex  ciphers  to  tJie  remainder,  and  continue 
dividing  until  there  are  3  or  4  decimal  places  in  the  quo- 
tient, and  then  if  there  still  be  a  remainder,  terminate  t/u 
qiwtient  by  the  plus  sign  (+),  which  shows  that  it  can  be 
carried  further. 


1  ARITHMETIC.  45 

161.      EXAMPLE. — What  is  the  quotient  of  199  divided  by  15  ? 

divisor  dividend  quotient 

SOLUTION.—  15)199(13  +  ^    Ans. 

1  5 
49 
45 
remainder    4 

Or,     1  5  )  1  9  9.0  0  0  (  1  3.2  6  6  +    Ans. 
15 

49 
45 


40 
30 
100 
9_0_ 

To~o 
£1 

remainder    1  0 
13T\  =  13.266  + 
£  =      .266  + 

162.  It  frequently  happens,  as  in  the  above  example, 
that  the  division  will  never  terminate.  In  such  cases,  decide 
to  how  many  decimal  places  the  division  is  to  be  carried, 
and  carry  the  work  one  place  further.  If  the  last  figure  of 
the  quotient  thus  obtained  is  5  or  a  greater  number,  increase 
the  preceding  figure  by  1,  and  write  after  it  the  minus  sign 
(— ),  thus  indicating  that  the  quotient  is  not  quite  as  large 
as  indicated ;  if  the  figure  thus  obtained  is  less  than  5,  write 
the  plus  sign  (-{-)  after  the  quotient,  thus  indicating  that 
the  number  is  slightly  greater  than  as  indicated.  In  the 
last  example,  had  it  been  desired  to  obtain  the  answer  cor- 
rect to  four  decimal  places,  the  work  would  have  been  car- 
ried to  five  places,  obtaining  13.26666,  and  the  answer  would 
have  been  given  as  13.2667  —  .  This  remark  applies  to  any 
othej-  calculation  involving  decimals,  when  it  is  desired  to 
omit  some  of  the  figures  in  the  decimal.  Thus,  if  it  was  de- 
sired to  retain  three  decimal  places  in  the  number  .2471253, 
it  would  be  expressed  as  .  247  + ;  if  it  was  desired  to  retain 
five  decimal  places,  it  would  be  expressed  as  .24713  —  . 
Both  the  -f-  and  —  signs  are  frequently  omitted;  they  are 


46 


ARITHMETIC. 


seldom  used  cmtside  of  arithmetic,  except  in  exact  calcula- 
tions, when  it  is  desired  to  call  particular  attention  to  the 
fact  that  the  result  obtained  is  not  quite  exact. 


EXAMPLES  FOR  PRACTICE. 


163.      Divide: 

(a)  101.6688  by  2.36. 

(b)  187.12264  by  123.107. 

(c)  .08  by  .008. 
.0003  by  3.75. 
.0144  by  .024. 
.00375  by  1.25. 
.004  by  400. 

.4  by  .008. 


(d) 
w 


Ans. 


(a)  43.08. 

(&)  1.52. 

(c)  10. 

(d)  .00008. 
(*)  .6. 
(/)  .003. 
(g)  .00001. 
(//)  .  50. 


REDUCTION    OF    DECIMATES. 


TO  REDUCE  A  FRACTION  TO  A  DECIMAL. 

164.      EXAMPLE. —    f  equals  what  decimal  ? 
SOLUTION.—  4  )  3.0  0 


.75 

EXAMPLE.— What  decimal  is  equivalent  to  £  ? 
SOLUTION.—  8)  7.0  00  (.8  7  5 

64 


Ans. 


=  .875.     Ans. 


40 

£0 

0 


165.  Rule. — Annex  ciphers  to  the  numerator  and  divide 
by  the  denominator.  Point  off  as  many  decimal  places  in  the 
quotient  as  there  are  ciphers  annexed. 


166. 


EXAMPLES  FOR  PRACTICE. 

Reduce  the  following  common  fractions  to  decimals: 


(a] 
(*) 


H 
H- 
A 
I 

¥T 

1000' 


Ans. 


W 


.46875. 

.875. 

.65625. 

.796875. 

.16. 

.625. 

.05. 

.004. 


§  1  ARITHMETIC.  47 

167.  To  reduce  indies  to  decimal  parts  of  a  foot: 

EXAMPLE. — What  decimal  part  of  a  foot  is  9  inches  ? 

SOLUTION. — Since  there  are  12  inches  in  one  foot,  1  inch  is  ^  of  a 
foot,  and  9  inches  is  9  X  r^>  °r  -f?  of  a  foot.  This  reduced  to  a  decimal 
by  the  above  rule  shows  what  decimal  part  of  a  foot  9  inches  is. 

1  2  )  9.0  0  ( .7  5  of  a  foot.     Ans. 
84 
60 
6_0 
0 

168.  Rule. — I.      To  reduce  inches  to  a  decimal  part  of 
a  foot,  divide  the  number  of  inches  by  12. 

II.  Should  the  resulting  decimal  be  an  unending  one,  and 
it  is  desired  to  terminate  the  division  at  some  point,  say  the 
fourth  decimal  place,  carry  the  division  one  place  further, 
and  if  the  fifth  -figure  is  5  or  greater  increase  the  fourth 
figure  by  1,  omitting  the  signs  -f-  and  — . 


(*) 

.25  ft. 

(^) 

.375  ft. 

Ans.  « 

w 

.4167  ft. 

(rtT) 

.5521  ft. 

w 

.9167  ft. 

EXAMPLES  FOR  PRACTICE. 

160.      Reduce  to  the  decimal  part  of  a  foot: 
(a)     3  in. 
(£)     4£  in. 

(c)  5  in. 

(d)  6f  in. 

(e)  11  in. 


TO  REDUCE  A  DECIMAL  TO  A  FRACTION. 

1 7O.  EXAMPLE. — Reduce  .125  to  a  fraction. 
SOLUTION.—    .125  =  ^ffo  =  ^  =  £.     Ans. 
EXAMPLE. — Reduce  .875  to  a  fraction. 
SOLUTION.—    .875  =  /ow  =  ft  =  I-     Ans. 

• 

171.  Rule. —  Under  the  figures  of  the  decimal,  place  1 

with  as  many  ciphers  at  its  right  as  there  are  decimal  places 
in  the  decimal,  and  reduce  the  resulting  fraction  to  its  lozvest 
terms  by  dividing  both  numerator  and  denominator  by  the 
same  number. 


48 


ARITHMETIC. 


172. 


EXAMPLES  FOR  PRACTICE. 

Reduce  the  following  to  common  fractions: 


(a) 


.125. 

.625. 

.3125. 

.04. 
(e)      .06. 
(/)    -75. 
(£-)    .15625. 
(A)     .875. 


(d) 


Ans. 


(«) 
(*) 
(') 


f 


(')  iV 

(/)  f 

teO  A" 

(*)  f 


173.  To  express  a  decimal  approximatively  as  a 
fraction  having  a  given  denominator: 

174.  EXAMPLE.— Express  .5827  in  64ths. 

SOLUTION.—     .5827  X  f£  =  87'ff28,  say  f|. 

o4 

Hence,  .5827  =  f|,  nearly.     Ans. 

EXAMPLE.— Express  .3917  in  12ths. 

4  7004 
SOLUTION.—      .3917  X  if  =    '  2    ,  say  T5¥. 

Hence,  .3917  =  T5^,  nearly.     Ans. 

175.  Rule. — Reduce  1  to  a  fraction  having  the  given 
denominator.     Multiply  the  given  decimal  by  the  fraction  so 
obtained,  and  the^result  will  be  the  fraction  required. 


EXAMPLES  FOR  PRACTICE. 


176.      Express: 


(*) 


M 


.625  in  8ths. 
.3125  in  16ths. 
.15625  in  32ds. 
.77in64ths. 
.81  in  48ths. 
.923  in  96ths. 


Ans.  • 


(a) 
(*) 

(d) 


ft 


177.  The  sign  for  dollars  is  $.  It  is  read  dollars.  125 
is  read  25  dollars. 

Since  there  are  100  cents  in  a  dollar,  1  cent  is  1  one-hun- 
dredth of  a  dollar ;  the  first  two  figures  of  a  decimal  part  of 


g  1  ARITHMETIC.  49 

a  dollar  represent  cents.     Since  a  mill  is  y1^  of  a  cent,  or 
TTJ?T5.Tr  of  a  dollar,  the  third  figure  represents  mills. 

Thus,  $25.16  is  read  twenty-Jive  dollars  and -six  teen  cents; 
$25.168  is  read  tzventy-five  dollars  sixteen  cents  and  eight 
mills. 


SYMBOLS  OF  AGGREGATION. 

178.  The  vinculum ,  parenthesis  ( ),  brackets  [  ], 

and  brace  { }  are  called  symbols  of  aggregation,  and  are 
used  to  include  numbers  which  are  to  be  considered  together ; 
thus,  13X8  —  3,  or  13  X  (8  —  3),  shows  that  3  is  to  be  taken 
from  8  before  multiplying  by  13. 

13x(8-3)  =  13X5  =  65. 
13X8^  =  13X5  =  65. 

When  the  vinculum  or  parenthesis  is  not  used,  we  have 
13X8-3  =  104-3  =  101. 

179.  In  any  series  of  numbers  connected  by  the  signs 
-f-,  — ,  X,   and  -T-,  the  operations  indicated  by   the  signs 
must  be  performed  in  order  from  left  to  right,  except  that 
no  addition  or  subtraction  may  be  performed  if  a  sign  of 
multiplication  or  division  follows  the  number  on  the  rigJit 
of  a  sign  of   addition    or   subtraction    until   the   indicated 
multiplication  or  division  has  been  performed.     In  all  cases 
the  sign  of  multiplication  takes  the  precedence,  the  reason 
being  that  when  two  or  more  numbers  or  expressions  are 
connected  by  the  sign  of  multiplication  the  numbers  thus 
connected  are  regarded  as  factors  of  the  product  indicated, 
and  not  as  separate  numbers. 

EXAMPLE.— What  is  the  value  of  4  X  24  —  8  +  17? 
SOLUTION. — Performing  the  operations  in  order  from  left  to  right, 
4X24  =  96;  96-8  =  88;  88  +  17  =  105.     Ans. 

180.  EXAMPLE. — What  is  the  value  of  the  following  expression: 
1,296  -4- 12  +  160  -  22  X  3£  =  ? 

SOLUTION.—  1,296-*- 12  =  108;  108  +  160  =  268;  here  we  cannot  sub- 
tract 22  from  268  because  the  sign  of  multiplication  follows  22 ;  hence, 
multiplying  22  by  3*,  we  get  77,  and  268  -  77  =  191.  Ans. 


50  ARITHMETIC.  §  1 

Had  the  above  expression  been  written  1,296-^-12  +  160 

—  22  X  3£  -4-  7  -f-  25,    it  would  have   been  necessary  to  have 
divided  22  X  3^  by  7  before  subtracting,  and  the  final  result 
would  have  been  22  X3£  =  77;  77-*- 7  =  11;  268-11  =  257; 
257  +  25  =  282.      Ans.      In  other  words,  it  is  necessary  to 
perform  all  the  indicated  multiplication  or  division  included 
between  the  signs  -j-  and  — ,  or  —  and  -j- ,  before  adding  or 
subtracting.     Also,  had  the  expression  been  written  1,296 
-T-  12  +  160  —  24|  -+  7  X  3|  +  25,  it  would  have  been  necessary 
to  have  multiplied  3^  by  7  before  dividing  24-J-,  since  the 
sign  of  multiplication  takes  the  precedence,  and  the  final 
result   would  have  been  3|X7  =  24|;  24|-+24|  =  1;    268 

—  1  =  267;  267  +  25  =  292.     Ans. 

It  likewise  follows  that  if  a  succession  of  multiplication 
and  division  signs  occur,  the  indicated  operations  must  not 
be  performed  in  order,  from  left  to  right — the  multiplication 
must  be  performed  first.  Thus,  24x3-=-4x2-+9x5  =  £. 
Ans.  In  order  to  obtain  the  same  result  that  would  be 
obtained  by  performing  the  indicated  operations  in  order, 
from  left  to  right,  symbols  of  aggregation  must  be  iised. 
Thus,  by  using  two  vinculums  the  last  expression  becomes 
24x3-f-4x2-=-9x5  =  20,  the  same  result  that  would  be 
obtained  by  performing  the  indicated  operations  in  order, 
from  left  to  right. 


EXAMPLES  VOR  PRACTICE. 

181.      Find  the  values  of  the  following  expressions : 


(a) 

(b)  5X24-32. 

(c)  5  X  24  -*- 15. 
(tt)  144  _  5  X  24. 

(e)  (1,691 -540 +  559)-- 3X57.  Ans> 

(/)  2,080  +  120-80X4-1,670. 

(g}  (90  +  60  -r-  25)  X  5  -  29. 

(//)  90  +  60  -s-  25X5. 


(a)  3. 

(b)  88. 
(0  8. 

(d)  24. 

(e)  10. 
(/)  210. 


1.2. 


ARITHMETIC. 

(CONTINUED.) 

PERCENTAGE. 

1.  Percentage,  is  the  process  of  calculating  by  hun- 
dredths. 

2.  The  term  per  cent,  is  an  abbreviation  of  the  Latin 
words  per  centum,  which  mean  by  the  hundred.     A  certain 
per  cent,  of  a  number  is  the  number  of  hundredths  of  that 
number  which  is  indicated  by  the  number  of  units  in  the 
percent.     Thus,  6  per  cent,   of  126  is  125  X^Q  =  7.5;  25 
per  cent,  of  80  is  80  X  T2^-  =  20 ;  43  per  cent,  of  432  pounds  fs 
432XiV\r  =  185.76  pounds. 

3.  The  sign  of  per  cent,    is  fa    and  is  read  per  cent. 
Thus,  6$  is  read  six  per  cent.;  V&\%  is  read  twelve  and  one- 
half  per  cent.,  etc. 

When  expressing  the  per  cent,  of  a  number  to  use  in  cal- 
culations, it  is  customary  to  express  it  decimally  instead  of 
fractionally.  Thus,  instead  of  expressing  Qfa  25$,  and  43$ 
as  yJ7,  T2^,  and  y4^,  it  is  usual  to  express  them  as  .06,  .25, 
and  .43. 

The  following  table  will  show  how  any  per  cent,  can  be 
expressed  either  as  a  decimal  or  as  a  fraction : 


Per  Cent. 

Decimal. 

Fraction. 

Per  Cent. 

Decimal. 

Fraction. 

1$ 

01 

T^TT 

150$ 

1  50 

44£  or  14 

2% 

02 

ITR) 

2    or—TTf 

500$ 

5  00 

TTTTT  or  5 

5$ 

05 

j>    or  -rV 

1$ 

0025 

\__  or  T-irj 

10$ 

10 

10    or    l 

1$  . 

.005  . 

i_.  or  -sijc 

25$ 

.25 

25     or     1 

u$ 

.015 

Ji_  or  •*£  A 

50$.      . 

.50 

T5n°n  or    i 

8i$.. 

.081 

2L  or  JL 

75$  

.75 

T7A  or  | 

12^$  

.125 

I  (TIT            12 

-^Eor    i 

100$ 

1  00 

4£S  or  1 

168% 

16| 

THO             » 

16i  or    i 

125<£ 

1  25 

*•%*  or  14 

624$ 

625 

Ttffr            ^ 

62ior    | 

iTJff  Wi    A¥ 

w-g/v.  . 

Tirti         * 

2  ARITHMETIC.  §  8 

4.  The  names  of  the  different  elements  used  in  percent- 
age are:  the  base,  the  rate  per  cent.,   the  percentage,   the 
amount,  and  the  difference. 

5.  The  base  is  the  number  on  which  the  per  cent,  is 
comptited. 

6.  The  rate  is  the  number  of  hundredths  of  the  base  to 
be  taken. 

7.  The   percentage   is   the   part,   or  number   of   him- 
drcdtlis,  of  the  base  indicated  by  the  rate ;  or,  the  percentage 
is  the  result  obtained  by  multiplying  the  base  by  the  rate. 

Thus,  when  it  is  stated  that  7#  of  $25  is  $1.75,  $25  is  the 
base,  7#  is  the  rate,  and  $1. 75  is  the  percentage. 

8.  The  amount  is  the  sum  of  the  base  and  percentage. 

9.  The  difference  is  the  remainder  obtained  by  sub- 
tracting the  percentage  from  the  base. 

Thus,  if  a  man  has  $180,  and  he  earns  6$  more,  he  will  have 
altogether  $180  +  $180  X.  06,  or  $180 +  $10. 80  =  $190.80. 
Here  $180  is  the  base;  6#,  the  rate;  $10.80,  the  percentage; 
and  $190.80,  the  amount. 

Again,  if  an  engine  of  125  horsepower  uses  16$  of  it  in 
overcoming  friction  and  other  resistances,  the  amount  left 
for  obtaining  useful  work  is  125  — 125  X.I  6  =  125  —  20  =  105 
horsepower.  Here  125  is  the  base;  16$,  the  rate;  20,  the 
percentage;  and  105,  the  difference. 

10.  From  the  foregoing  it  is  evident  that  to  find  the 
percentage,  the  base  must  be  multiplied  by  the  rate.     Hence, 
the  following 

Rule. —  To  find  the  percentage,  multiply  the  base  by  the 
rate  expressed  decimally. 

EXAMPLE. — Out  of  a  lot  of  300  bushels  of  apples  76#  were  sold.  How 
many  bushels  were  sold  ? 

SOLUTION. —  76$,  the  rate,  expressed  decimally,  is  .76;  the  base  is 
300;  hence,  the  number  of  bushels  sold,  or  the  percentage,  is,  by  the 

above  rule, 

300  X  .76  =  228  bushels.     Ans. 

Expressing  the  rule  as  a 

Formula,  percentage  =  base^rate. 


§  2  ARITHMETIC.  3 

11.  When  the  percentage  and  rate  are  given,  the  base 
may  be  found  by  dividing  the  percentage  by  the  rate.     For, 
suppose  that  12  is  G#,  or  -j-J^,  of  some  number;  then  \%,  or 
•j-J^,  of  the  number,  is  12  -i-  6,  or  2.     Consequently,  if  2  .=  Itf, 
or  y^,  100#,  or  |££  =  2  X 100  =  200.     But,  since  the  same 
result  may  be  arrived  at  by  dividing  12  by  .06,  for  12-f-.06 
=  200,  it  follows  that . 

Rule. —  When  the  percentage  and  rate  are  given,  to  find  the 
base,  divide  the  percentage  by  the  rate  expressed  decimally. 

Formula,  base  =  percentage  -4-  rate. 

EXAMPLE. — Bought  a  certain  number  of  bushels  of  apples  and  sold 
r  them.     If  I  sold  228  bushels,  how  many  bushels  did  I  buy  ? 

SOLUTION. — Here  228  is  the  percentage,  and  76£,  or  .76,  is  the  rate; 
hence,  applying  the  rule, 

228  -f-  .76  =  300  bushels.     Ans. 

12.  When  the  base  and  percentage  are  given,  to  find  the 
rate,  the  rate  may  be  found,  expressed  decimally,  by  divi- 
ding the  percentage  by  the  base.     For,  suppose  that  it  is 
desired  to  find  what  per  cent.   12  is  of  200.     1#  of  200  is 
•^«>0  x  .01  =  2.     Now,  if  \%  is  2,  12  is  evidently  as  many  per 
cent,  as  the  number  of  times  that  2  is  contained  in  12,  or 
1 2  -j-  2  = .  6^.     But  the   same  result   may  be   obtained  by 
dividing  12,  the  percentage,  by  200,  the  base,  since  12 -f- 200 
=  .0fj  =  6#.     Hence, 

Rule. —  When  the  percentage  and  base  are  given,  to  find 
the  rate,  divide  the  percentage  by  the  base,  and  the  result  will 
be  the  rate  expressed  decimally. 

Formula,  rate  =  percentage  -5-  base. 

E  x  AMPLE.— Bought  300  bushels  of  apples  and  sold  228  bushels.  What 
per  cent,  of  the  total  number  of  bushels  was  sold  ? 

SOLUTION. — Here  300  is  the  base  and  228  is  the  percentage;  hence, 
applying  rule,  rate  =  228-^300  =  .76  =  76*.  Ans. 

EXAMPLE.— What  per  cent  of  875  is  25  ? 

SOLUTION. — Here  875  is  the  base,  and  25  is  the  percentage;  hence, 
applying  rule,  25 -•- 875  =  .OSf  =  2f*.  Ans. 

PROOF.—    875  X  .<&f  =  25. 


ARITHMETIC.  §  2 


EXAMPLES    FOR    PRACTICE. 
13.      What  per  cent,  of: 


(a) 
(*) 

(') 

360 
900 
125 
150 

is 
is 
is 
is 

90? 
360? 

25? 
750? 

(*) 

280 

is 

112? 

(/) 

400 

is 

200? 

te) 

47  is  94  ? 
500  is  250  ? 

(e)  40#. 
(/)  50#. 
(A'')  200#. 


14.  The  amount  may  be  found,   when   the   base  and 
rate  are  given,  by  multiplying-  the  base  by  1  plus  the  rate, 
expreSvSed  decimally.      For,  suppose  that  it  is  desired  to  find 
the  amount  when  200  is  the  base  and  G$  is  the  rate.     The 
percentage  is  200  X-  06  =  12,  and,  according  to  definition, 
Art.  8,  the  amount  is  200  +  12  =  212.     But,  the  same  result 
may  be  obtained  by  multiplying  200  by  1-f.OG,  or   LOG, 
since  200  X  LOG  =  212.     Hence, 

Rule.  —  When  the  base  and  rate  are  given,  to  find  the 
amount,  multiply  the  base  by  1  plus  the  rate  expressed 
decimally. 

Formula,  amount  =  basexQ  -\-ratc). 

EXAMPLE.  —  If  a  man  earned  $725  in  a  year,  and  the  next  year  10£ 
more,  how  much  did  he  earn  the  second  year  ? 

SOLUTION.  —  Here  725  is  the  base  and  10$  is  the  rate,  and  the  amount 
is  required.  Hence,  applying  the  rule, 

725X1.10  =  $797.50.     Ans. 

15.  When  the  base  and  rate  are  given,  the  difference 
may  be  found  by  multiplying  the  base  by  1  minus  the  rate 
expressed  decimally.     For,  suppose  that  it  is  desired  to  find 
the  difference  when  the  base  is  200  and  the  rate  is  6#.     The 
percentage  is  200  X.  06  =  12;  and,  according  to  definition, 
Art   9,  the   difference  =  200  —  12  =  188.     But,  the   same 
result  may  be  obtained  by  multiplying  200  by  1  —  .06,  or  .94, 
since  200  X.  94  =  188.     Hence, 

Rule.  —  When  the  base  and  rate  are  given,  to  Jin  d  the  differ- 
ence, multiply  the  base  by  1  minus  the  rate  expressed  decimally. 
Formula,  difference  =  base  X(L  —  rate).  ' 


§  2  ARITHMETIC.  5 

EXAMPLE. — Bought  300  bushels  of  apples  and  sold  all  but  24$  of 
them.  How  many  bushels  were  sold  ? 

SOLUTION. — Here  300  is  the  base,  24$  is  the  rate,  and  it  is  desired  to 
find  the  difference.  Hence,  applying  the  rule, 

300  X  (1  -  .24)  =  228  bushels.     Ans. 

16.  When  the  amount  and  rate  are  given,  the  base  may 
be  found  by  dividing  the  amount  by  1  plus  the  rate.     For, 
suppose   that   it   is  known    that   212   equals  some  number 
increased  by  G$  of  itself.     Then,  it  is  evident  that  212  equals 
106$  of  the  number  (base)  that  it  is  desired  to  find.     Conse- 

010 

quently,  if  212  =  10G$,  10  =  —  =  2,  and  100$  =  2x100 

=  200  =  the  base.  But  the  same  result  may  be  obtained 
by  dividing  212  by  l  +  .OG,  or  LOG,  since  212 4- LOG  =  200. 
Hence, 

Rule. —  When  the  amount  and  rate  are  given^  to  find  the 
base,  divide  the  amount  by  1  plus  the  rate  expressed  decimally. 
Formula,  base  =  amount  -f-  (1  +  rate). 

EXAMPLE. — The  theoretical  discharge  of  a  certain  pump  when  run- 
ning at  a  piston  speed  of  100  feet  per  minute  is  278,910  gallons  per  day 
of  10  hours.  Owing  to  leakage  and  other  defects,  this  value  is  25$ 
greater  than  the  actual  discharge.  What  is  the  actual  discharge  ? 

SOLUTION. — Here  278,910  equals  the  actual  discharge  (base)  increased 
by  25$  of  itself.  Consequently,  278,910  is  the  amount,  and  25$  is  the 
rate.  Applying  rule, 

actual  discharge  =  278,910  -*•  1.25  =  223,128  gallons.     Ans. 

17.  When  the  difference  and  rate  are  given,  the  base 
may  be  found  by  dividing  the  difference  by  1  minus  the  rate. 
For,  suppose  that  188  equals  some  number  less  G$  of  itself. 
Then,  188  evidently  equals  100  —  6  =  94$  of  some  number. 
Consequently,  if  188  =  94$,  1$  =  188-^-94  =  2,   and  100$ 
=  2x100  =  200.     But  the  same  result  may  be  obtained  by 
dividing  188  by  1  — .  OG,  or  .  94,  since  188  -f- .  94  =  200.   Hence, 

Rule. —  When  the  difference  and  rate  are  given,  to  find  the 
base,  divide  the  difference  by  1  minus  the  rate  expressed 
decimally. 

Formula,  base  —  difference  -4-  (1  —  rate). 

1-5 


6  ARITHMETIC.  §  2 

EXAMPLE. — Bought  a  certain  number  of  bushels  of  apples  and  sold 
76$  of  them.  If  there  were  72  bushels  left  unsold,  how  many  bushels 
did  I  buy  ? 

SOLUTION. — Here  72  is  the  difference  and  76$  is  the  rate.  Applying 
rule,  72  -*-  (1  -  .76)  ==  300  bushels.  Ans.  ' 

EXAMPLE. — The  theoretical  number  of  foot-pounds  of  work  per  min- 
ute required  to  operate  a  boiler  feed-pump  is  127,344.  If  30$  of  the 
total  number  actually  required  be  allowed  for  friction,  leakage,  etc. , 
how  many  foot-pounds  are  actually  required  to  work  the  pump  ? 

SOLUTION. — Here  the  number  actually  required  is  the  base;  hence, 
127,344  is  the  difference,  and  30$  is  the  rate.     Applying  the  rule, 
127,344  -T-  (1  -  .30)  =  181,920  foot-pounds.     Ans. 

18.  EXAMPLE. — A  certain  chimney  gives  a  draft  of  2.76  inches  of 
water.     By  increasing  the  height  20"  feet,  the  draft  was  increased  to  3 
inches  of  water.     What  was  the  gain  per  cent.  ? 

SOLUTION. — Here  it  is  evident  that  3  inches  is  the  amount,  and  that 
2.76  inches  is  the  base.  Consequently,  3  —  2.76  =  .24  inch  is  the  per- 
centage, and  it  is  required  to  find  the  rate.  Hence,  applying  the  rule 
given  in  Art.  12, 

gain  per  cent.  =  .24-4-2.76  =  .087  =  8.7$.     Ans. 

19.  EXAMPLE. — A  certain  chimney  gave  a  draft  of  3  inches  of 
water.     After  an  economizer  had  been  put  in,  the  draft  was  reduced  to 
1.2  inches  of  water.     What  was  the  loss  per  cent.? 

SOLUTION. — Here  it  is  evident  that  1.2  inches  is  the  difference  (since 
it  equals  3  inches  diminished  by  a  certain  per  cent,  loss  of  itself),  and 
3  inches  is  the  base.  Consequently,  3  —  1.2  =  1.8  inches  is  the  percent- 
age. Hence,  applying  the  rule  given  in  Art.  12, 

loss  per  cent.  =  1.8  -*-  3  =  .60  =  60$.     Ans. 

20.  To  flntl  the  gain  or  loss  per  cent.  : 

Rule. — Find  the  difference  between  the  initial  and  the  final 
value;  divide  this  difference  by  the  initial  value. 

EXAMPLE. — If  a  man  buys  a  house  for  $1,860,  and  some  time  after- 
wards builds  a  barn  for  25$  of  the  cost  of  the  house,  does  he  gain  or 
lose,  and  how  much  per  cent,  if  he  sells  both  house  and  barn  for 
$2,100? 

SOLUTION.— The  cost  of  the  barn-was  $1,860  X-25  =  $465;  conse- 
quently, the  initial  value,  or  total  cost,  was  $1,860 +  $465  =  $2,325. 
Since  he  sold  them  for  $2,100  he  lost  $2,325  — $2,100  =  $225.  Hence, 
applying  rule, 

225-^-2,325  =  .0968  =  9. 68$  loss.     Ans. 


(a)     $112.50. 
£       5.016. 


940.8. 


§  2  ARITHMETIC. 

EXAMPLES  FOB  PRACTICE. 

J31«      Solve  the  following  : 
(a)     What  is  12|$  of  $900  ? 
(£)      What  is  |$  of  627  ? 
(;•)      What  is  33£$  of  54  ? 
(a?)     101  is  68f  $  of  what  number  ? 
(e)     784  is  83i$  of  what  number  ? 
(/)    What  $  of  960  is  160  ? 
(g)    What  $  of  $3,606  is  $450f  ? 
(>&)     What  $  of  280  is  112  ? 

1.  A  steam  plant  consumed  an  average  of  3,640  pounds  of  coal  per 
day.     The  engineer  made  certain  alterations  which  resulted  in  a  saving 
of  250  pounds  per  day.     What  was  the  per  cent,  of  coal  saved  ? 

Ans.  7$,  nearly. 

2.  If  the  speed  of  an  engine  rtrnning  at  126  revolutions  per  minute 
should  be  increased  6|$,  how  many  revolutions  per  minute  would  it 
then  make  ?  Ans.  134.19  rev. 

3.  The  list  price  of  a  lot  of  silk  goods  is  $1,400,  of  some  laces  $1,150, 
and  of  some  calico  $340.     If  25$  discount  was  allowed  on  the  silk,  22$ 
on  the  laces,  and  12^$  on  the  calico,  what  was  the  actual  cost  of  the 
purchase?  Ans.  $2,244.50. 

4.  If  I  loan  a  man  $1,100,  and  this  is  18|$  of  the  amount  that  I  have 
on  interest,  how  much  money  have  I  on  interest  ?  Ans.  $5,945.95. 

5.  A  test  showed  that  an  engine  developed  190.4  horsepower,  15$ 
of  which  was  consumed  in  friction.     How  much  power  was  available 
for  use?  Ans.  161.84  H.  P. 

6.  By  adding  a  condenser  to  a  steam  engine,  the  power  was  increased 
14$  and  the  consumption  of  coal  per  horsepower  per  hour  was  decreased 
20$.     If  the  engine  could  originally  develop  50  horsepower,  and  required 
3^  pounds  of  coal  per  horsepower  per  hour,  what  would  be  the  total 
weight  of  coal  used  in  an  hour,  with  the  condenser,  assuming  the  engine 
to  run  full  power  ?  Ans.  159.6  pounds. 


DE3STOMIKATE  LUMBERS. 

22.  A  denominate  number  is  a  concrete  number,  and 
may  be  either  simple  or  compotind;  as,  8  quarts;  5  feet;  10 
inches,  etc. 

23.  A  simple  denominate  number  consists  of  units 
of  but  one  denomination;  as,  1C  cents;   10  hours;  5  dollars, 
etc. 


8  ARITHMETIC.  §  2 

24.  A   compound  denominate  number   consists   of 
units  of  two  or  more  denominations  of  a  similar  kind ;  as,  3 
yards,  2  feet,  1  inch ;  34  square  feet,  57  square  inches. 

25.  In  whole  numbers  and  in  decimals  the  law  of 

increase  and  decrease  is  on  the  scale  of  10,  but  in  com- 
pound or  denominate  numbers  the  scale  varies. 

26.  A  measure  is  a  standard  unit,  established  by  law 
or  custom,  by  which  quantity  of  any  kind  is  measured.     The 
standard  unit  of  dry  measure  is  the  Winchester  bushel ;  of 
weight,  the  pound ;  of  liquid  measure,  the  gallon,  etc. 

27.  Measures  are  of  six  kinds : 

1.  Extension.  4.     Time. 

2.  Weight.  5.     Angles. 

3.  Capacity.  6.  .   Money  or  value. 


MEASURES  OF  EXTENSION. 

28.     Measures   of  extension  are   used   in   measuring 
lengths,  distances,  surfaces,  and  solids. 

LINEAR  MEASURE. 


TABLE. 
Abbreviation. 


12    inches  (in.)  -  1  foot .    .     ft. 

3    feet  ...     =1  yard     .  yd. 

5. 5  yards    .    .    =  1  rod  .    .    rd. 
40    rods  .    .    .    =  1  furlong  fur. 

8    furlongs    .    =  1  mile     .  mi. 


in.  ft.  yd.        rd.  fur.  mi. 

36=    3       =1 
198  =  16|     =5.5     =      1 
7,920  =  660     =220     =40  =  1 
63,360  =  5,280  =  1,760  =  320  =  8  =  1 


SURVEYOR'S  LINEAR  MEASURE. 

TABLE. 

7.92  inches  =  1  link li. 

25  links  =  1  rod rd. 

4  rods    ) 
100  links  [  1  clmm     •••-<&• 

80  chains  =  1  mile mi. 

mi.         ch.  rd.  li.  in. 

1    =    80    =    320    =    8,000    =    63,360 

29.     The  linear  unit,   generally  used  by  surveyors,   is 
Gunter's  chain,  which  is  equal  to  4  rods,  or  66  feet. 


§  2  ARITHMETIC.  9 

3O.  An  engineer's  chain,  used  by  civil  engineers,  is 
100  feet  long,  and  consists  of  100  links.  In  computations, 
the  links  are  written  as  so  many  hundredths  of  a  chain. 


SQUARE  MEASURE. 

TABLE. 

144    square  inches  (sq.  in.)     .     .     . 

9    square  feet •  . 

30£  square  yards 


160 
640 


square  rods 
acres 


sq.  mi.       A.          sq.  rd.          sq.  yd. 


1  square  foot  ....  sq.  ft. 
1  square  yard  .  .  .  .  sq.  yd. 
1  square  rod  .  .  .  .  sq.  rd. 

1  acre A. 

1  square  mile  ....  sq.  mi. 
sq.  ft.  sq.  in. 


1      =  640  =  102,400  =  3,097,600  =  27,878,400  =  4,014,489,600 


SURVEYOR'S  SQUARE  MEASURE. 

TABLE. 

625  square  links  (sq.  li.)     .     .     .     .  =     1  square  rod     .  . 

16  square  rods =     1  square  chain 

10  square  chains =     1  acre      .     .     .  . 

640  acres =    1  square  mile  .  . 

36  square  miles  (6  mi.  square)  .     .  =     1  township  .     .  . 
sq.  mi.       A.        sq.  ch.       sq.  rd.           sq.  li. 
1      =  640  =  6,400  =  102,400  =  64,000,000 


sq.  rd. 
sq.  ch. 

.  A. 
sq.  mi. 

.  Tp. 


CUBIC  MEASURE. 

TABLE. 

.  =     1.  cubic  foot 
,     .  =     1  cubic  yard 


1,728    cubic  inches  (cu.  in.)    .     . 

27    cubic  feet 

128  cubic  feet =  1  cord 

24f  cubic  feet =  1  perch 

cu.  yd.     cu.  ft.       cu.  in. 
1       =    27    =    46,656 


cu.  ft. 

cu.  yd. 

.     cd. 

P. 


MEASURES  OF  WEIGHT. 

AVOIRDUPOIS  WEIGHT. 

TABLE. 

=    1  pound  .     . 


16  ounces  (oz.) 

100  pounds  .........  =     1  hundredweight 

20cwt.,  or  2,000  Ib  ......  =    1  ton  .... 

T.        cwt.  Ib.  oz. 

1    =    20    =  2,000    =    32,000 


Ib. 

cwt. 

T. 


10  .     ARITHMETIC.  §2 

31.  The  ounce  is  divided   into  halves,    quarters,   etc. 
Avoirdupois  weight  is  used  for  weighing  coarse  and  heavy 
articles.     One  avoirdupois  pound  contains  7,000  grains. 

LONG  TON  TABLE. 

16  ounces =    1  pound Ib. 

112  pounds =     1  hundredweight   .     .     .      cwt. 

20cwt,  or  2,240  Ib. "     ......=     1  ton T. 

32.  In  all  the  calculations  throughout  this  and  the  fol- 
lowing  sections,   2,000   pounds   will  be   considered   1   ton, 
unless  the  long  ton  (2,24=0  pounds)  is  especially  mentioned. 

TROY  WEIGHT. 

TABLE. 

24  grains  (gr.) =  1  pennyweight  ....  pwt. 

20  pennyweights =  1  ounce oz. 

12  ounces  .' =  1  pound Ib. 

Ib.        oz.         pwt.  gr. 

1    =-.    12    =    240    =    5,760 

33.  Troy  weight  is"  iised  in  weighing  gold  and  silver- 
ware, jewels,  etc.     It  is  used  by  jewelers. 


MEASURES  OF  CAPACITY. 

LIQUID  MEASURE. 

TABLE. 

4    gills  (gi.) =     1  pint pt. 

2    pints ._=    1  quart qt. 

4    quarts =     1  gallon gal. 

81-J-  gallons —     1  barrel bbl. 

2    barrels,  or  63  gallons                  .  =     1  hogshead hhd. 

hhd.     bbl.       gal.         qt.  pt.             gi. 

1    =    2    =    63    =    252  =    504    =    2,016 

DRY  MEASURE. 

TABLE. 

2  pints  (pt.) =     1  quart    .......     qt. 

8  quarts =     1  peck pk. 

4  pecks =     1  bushel bu. 

bu.      pk.      qt.        pt. 
1  =  4  =  32  =  64 


ARITHMETIC. 


11 


MEASURE  OF  TIME. 

TABLE. 


60  seconds  (sec  ) 

.     .     .  —    1  minute      .... 

min. 

GO  minutes                . 

.     .     .  —     1  hour     

.     .    hr. 

24  hours                    .     .     . 

.  —    1  day 

.     .    da. 

.     .  —     1  week     

365  days        [ 

.     .     .  —    1  common  year    .     . 

yr. 

12  months  j 
366  days                           .     . 

.     .     .  —     1  leap  year. 

100  years     

.     .  —     1  century. 

NOTE.  —  It  is  customary 

to  consider  one  month  as  30  days. 

MEASURE 

OF  ANGLES  Oil  ARCS. 

60  seconds  (") 
60  minutes 
90  degrees 
360  degrees      . 


TABLE. 

.     .     .....=     1  minute ' , 

. =     1  degree °. 

=     1  right  angle  or  quadrant  |__. 

=     1  circle    . cir. 

1  cir.  =  360°  =  21,600'  =  1,296,000" 


10  mills  (m.' 
10  cents  . 
10  dimes 
10  dollars 


MEASURE  OF  MOKEY. 

UNITED    STATES   MONEY. 

TABLE. 

.  =     1  cent . 


E. 

1  = 


ct. 

.     .     .     .  =    1  dime d. 

.     .     .     .  =    1  dollar $. 

.     .     .     .  =    1  eagle E. 

$  d.  ct.  m. 

10  =  100  =  1,000  =  10,000 


MISCELLANEOUS  TABLE. 


12  things  are  1  dozen. 

12  dozen  are  1  gross. 

12  gross  are  1  great  gross. 

2  things  are  1  pair. 
20  things  are  1  score. 

1  league  is  3  miles. 

1  fathom  is  6  feet. 


1  meter  is  nearly  39.37  inches. 

1  hand  is  4  inches. 

1  palm  is  3  inches. 

1  span  is  9  inches. 
24  sheets  are  1  quire. 
20  quires,  or  480  sheets,  are  1  ream. 

1  bushel  contains  2,150.4  cubic  in. 
1  U.  S.  standard  gallon  (also  called  a  wine  gallon)  contains  231  cubic  in. 
1  U.  S.  standard  gallon  of  water  weighs  8.355  pounds,  nearly. 
1  cubic  foot  of  water  contains  7.481  U.  S.  standard  gallons,  nearly. 
1  British  imperial  gallon  weighs  10  pounds. 

It  will  be  of  great  advantage  to  the  student  to  carefully 
memorize  all  the  above  tables. 


12  ARITHMETIC.  §  2 

REDUCTION  OF  DENOMINATE  NUMBERS. 

34.  Reduction  of  denominate  numbers  is  the  process  of 
changing-  their  denomination  without  changing  their  value. 
They  may  be  changed  from  a  higher  to  a  lower  denomina- 
tion, or  from  a  lower  to  a  higher — either  is  reduction.     As 

2  hours  =  120  minutes. 
32  ounces  =  2  pounds. 

35.  Principle. — Denominate  numbers  are  changed  to 
lower  denominations  by  multiplying,  and  to  higher  denom- 
inations by  dividing. 

To   reduce  denominate   numbers   to  lower  denom- 
inations : 

36.  EXAMPLE. — Reduce  5  yd.  2  ft.  'fin.  to  inches. 
SOLUTION. —  yd.  ft.  in. 

527 
3 
15ft. 


17ft. 
1  2 
"34 

1  7 

2  0  4  in. 

7  in. 
211  inches.     Ans. 

EXPLANATION. — Since  there  are  3  feet  in  1  yard,  in  5  yards 
there  are  5 X3  or  15  feet,  and  15  feet  plus  2  feet  =  17  feet. 
There  are  12  inches  in  a  foot;  therefore,  12x17  =  204 
inches,  and  204  inches  plus  7  inches  =  211  inches  =  num- 
ber of  inches  in  5  yards  2  feet  and  7  inches. 

37.      EXAMPLE. — Reduce  6  hours  to  seconds. 
SOLUTION. —  6         hours. 

60 

360      minutes. 
60 


21600   seconds.     Ans. 


ARITHMETIC. 


13 


EXPLANATION. — As  there  are  60  minutes  in  1  hour,  in  6 
hours  there  are  6  X  60,  or  360,  minutes ;  as  there  are  no  min- 
utes to  add,  we  multiply  360  minutes  by  60,  to  get  the 
number  of  seconds. 

38.  In  order  to  avoid  mistakes,  if  any  denomination  be 
omitted,  represent  it  by  a  cipher.     Thus,  before  reducing"  3 
rods  6  inches  to  inches,  insert  a  cipher  for  yards  and  a  cipher 
for  feet,  as 

rd.      yd.      ft.      in. 
3006 

39.  Rule. — Multiply  the  number  representing  the  high- 
est denomination  by  the  number  of  units  in  the  next  loiver 
required  to  make  one  of  the  higher  denomination,  and  to  the 
product  add  the  number  of  given  units  of  that  lower  denomi- 
nation.    Proceed  in  this  manner  until  the  number  is  reduced 
to  the  required  denomination. 


EXAMPLES  FOB  PRACTICE. 


4O.  Reduce: 

(a)  4  rd.  2  yd.  2  ft.  to  ft. 

(b)  4  bu.  3  pk.  2  qt.  to  qt. 

(c)  13  rd.  5  yd.  2  ft.  to  ft. 

(d)  5  mi.  100  rd.  10  ft.  to  ft. 

(e)  8  Ib.  4  oz.  6  pwt.  to  gr. 
(/)  52  hhd.  24  gal.  1  pt.  to  pt. 
(g)  5  cir.  16°  20'  to  minutes. 
(h)  14  bu.  to  qt. 


Ans.  - 


(a) 


(If) 
w 


74ft. 
154  qt. 
231.5ft. 
28,060  ft. 
48,144  gr. 
26,401  pt. 
108,980'. 
448  qt. 


To  reduce  lower  to  higher  denominations  : 

41.      EXAMPLE. — Reduce  211  inches  to  higher  denominations. 
SOLUTION.—  12 )  2  1  1  in. 

3)  17ft. +7  in. 


5  yd.  +  2  ft.    Ans. 

EXPLANATION. — There  are  12  inches  in  1  foot;  therefore, 
211  divided  by  12  =  17  feet  and  7  inches  over.  There  are 
3  feet  in  1  yard;  therefore,  17  feet  divided  by  3  =  6  yards 


14  ARITHMETIC.  §  2 

and  2  feet  over.     The  last  quotient  and  the  two  remainders 
constitute  the  answer,  5  yards  2  feet  7  inches. 

42.     EXAMPLE. — Reduce    15,735    grains    Troy  weight   to   higher 
denominations. 

SOLUTION.—  24)15735  gr?  (655  pwt. 

.    144 
133 
1  20 


135 
120 

-      1  5  gr. 

20)655  pwt.  (  3  2  oz. 
60 
55 
40 

1  5  pwt. 

1  2  )  3  2  oz.  (  2  Ib. 
24 
8oz. 

EXPLANATION. — There  are  24  grains  in  1  pennyweight,  and 
in  15,735  grains  there  are  as  many  pennyweights  as  24  is  con- 
tained in  15,735,  or  655  pennyweights  and  15  grains  remain- 
ing. There  are  20  pennyweights  in  1  ounce,  and  in  655 
pennyweights  there  are  32  ounces  and  15  pennyweights 
remaining.  There  are  12  ounces  in  1  pound,  and  in  32 
ounces  there  are  2  pounds  and  8  ounces  remaining.  The 
last  quotient  and  the  three  remainders  constitute  the  answer, 
2  pounds  8  ounces  15  pennyweights  15  grains/ 

The  above  problem  is  worked  out  by  long  division,  because 
the  numbers  are  too  large  to  solve  easily  by  short  division. 
The  student  may  use  either  method. 

43.  Rule. — Divide  the  number  representing  the  denomi- 
nation given  by  the  number  of  units  of  this  denomination 
required  to  make  one  unit  of  the  next  higher  denomination. 
The  remainder  will  be  of  the  same  denomination,  but  the 
quotient  will  be  of  the  next  higher.  Divide  this  quotient  by 
the  number  of  units  of  its  denomination  required  to  make 
one  tinit  of  the  next  higlier.  Continue  until  the  highest 


ARITHMETIC. 


15 


denomination  is  reached,  or  until  there  is  not  enough  of  a 

denomination  left  to  make  one  of  the  next  higher.      The  last 
quotient  and  t/ie  remainders  constitute  tJie  required  result. 


EXAMPLES  FOB  PRACTICE. 

44.      Reduce  to  units  of  higher  denominations: 
(a)  7,460  sq.  in.;     (£)  7,580  sq.  yd.;     (c)  148,760  cu.  in.;     (<t)  7.896 
cu.  ft.  to  cd. ;    (e)  17,651";    (/)  1,120  cu.  ft.  to  cd. ;    (g)  8,000  gi. ;    (ti) 
36,450  Ib. 

(a)      5  sq.  yd.  6  sq.  ft.  116  sq.  in. 
(£)      1  A.  90  sq.  rd.  17  sq.  yd.  4  sq.  ft.  72  sq.  in. 
(c)      3  cu.  yd.  5  cu.  ft.  152  cu.  in. 
j  (d)    61  cd.  88  cu.  ft. 
'  {  (e)      4°  54'  11". 
(/)    8  cd.  96  cu.  ft. 
(g)    3  hhd.  61  gal. 
(k)      18  T.  4  cwt.  50  Ib. 


ADDITION  OF  DENOMINATE  NUMBERS. 

45.      EXAMPLE.— Find  the  sum  of  3  cwt.  46  Ib.  12  oz. ;   8  cwt.  12  Ib. 
13  oz. ;  12  cwt.  50  Ib.  13  oz. ;  27  Ib.  4  oz. 
SOLUTION. — 


T. 

cwt. 

Ib. 

oz. 

0 

3 

46 

12 

0 

8 

12 

13 

0 

12 

50 

13 

0 

0 

27 

4 

1  4  37  10    Ans. 

EXPLANATION.- — Begin  to  add  at  the  right-hand  column: 
4  + 13  +  13  +  12  =  42  ounces;  as  16  ounces  make  1  pound, 
42  ounces -r- 16  =  2  and  a  remainder  of  10  ounces,  or  2 
pounds  and  10  ounces.  Place  10  ounces  under  ounce  column 
and  add  2  pounds  to  the  next  or  pound  column.  Then, 
2  +  27  +  50  +  12  +  46  =  137  pounds;  as  100  pounds  make 
a  hundredweight,  137 -J- 100  =  1  .hundredweight  and  a 
remainder  of  37  pounds.  Place  the  37  under  the  pounds 
column,  and  add  1  hundredweight  to  the  next  or  hundred- 
weight column.  Next,  1  +  12  +  8  +  3  =  24  hundredweight. 


16  ARITHMETIC.  §  2 

20  hundredweight  make  a  ton;  therefore  24-4-20  =  1  ton 
and  4  hundredweight  remaining.  Hence,  the  sum  is  1  ton 
4  hundredweight  37  pounds  10  ounces.  Ans. 

46.      EXAMPLE.— What  is  the  sum  of  2  rd.  3  yd.  2  ft.  5  in. ;  6  rd.  1 
ft.  10  in. ;  17  rd.  11  in. ;  4  yd.  1  ft.? 

SOLUTION. — 


rd. 

yd. 

ft. 

in. 

2 

3 

2 

5 

6 

0 

1 

10 

17 

0 

0 

11 

0 

4 

1 

0 

26 

3| 

0 

.       2 

26 

3 

1 

8    Ans. 

or 

*  EXPLANATION. — The  sum  of  the  numbers  in  the  first 
column  =  26  inches,  or  2  feet  and  2  inches  remaining.  The 
sum  of  the  numbers  in  the  next  column  plus  2  feet  =  6 
feet,  or  2  yards  and  0  feet  remaining.  The  sum  of  the  next 
column  plus  2  yards  =  9  yards,  or  9  -i-  5-j-  =  1  rod  and  3^ 
yards  remaining.  The  sum  of  the  next  column  plus  1  rod 
=  26  rods.  To  avoid  fractions  in  the  sum,  the  -J  yard  is 
reduced  to  1  foot  and  6  inches,  which  added  to  26  rods 
3  yards  0  feet  and  2  inches  =  26  rods  3  yards  1  foot  8 
inches.  Ans. 

47.  EXAMPLE.— What  is  the  sum  of  47  ft.  and  3  rd.  2  yd.  2  ft. 
10  in.? 

SOLUTION. — When  47  ft.  is  reduced  it  equals  2  rd.  4  yd.  2  ft.  which 
can  be  added  to  3  rd.  2  yd.  2  ft.  10  in.     Thus, 

rd.  yd.  ft.  in. 

3  2  2  10 

2 4 2 0 

6  H  1  10 

or  6  2  0  4    Ans. 

48.  Rule. — Place  the  numbers  so  that  like  denominations 
are  under  each  other.     Begin  at  the  right-hand  column,  and 
add.     Divide  the  sum  by  the  number  of  units  of  this  denomi- 
nation required  to  make  one  unit  of  the  next  higher.     Place 
the   remainder   under    the    column   added,    and    carry    the 
quotient  to  the  next  column.      Continue  in  this  manner  until 
the  highest  denomination  given  is  reached. 


§  2  ARITHMETIC.  17 

EXAMPLES  FOB  PRACTICE. 

49.      What  is  the  sum  of: 

(a)  25  Ib.  7  oz.  15  pwt.  23  gr. ;  17  Ib.  16  pwt. ;  15  Ib.  4  oz.  12  pwt. ; 
18  Ib.  16  gr. ;  10  Ib.  2  oz.  11  pwt.  16  gr.? 

(b)  9  mi.  13  rd.  4  yd.  2  ft. ;  16  rd.  5  yd.  1  ft.  5  in. ;  16  mi.  2  rd.  3  in. ; 
14  rd.  1  yd.  9  in.  ? 

(c)  3  cwt.  46  Ib.  12  oz. ;  12  cwt.  9£  Ib. ;  2±  cwt  2l£  Ib.  ? 

(d)  10  yr.  8  mo.  5  wk.  3  da. ;  42  yr.  6  mo.  7  da. ;  7  yr.  5  mo.  18  wk. 
4  da.;  17  yr.  17  da.? 

(e)  17  T.  11  cwt.  49  Ib.  14  oz. ;  16  T.  47  Ib.  13  oz. ;  20  T.  13  cwt.  14 
Ib.  6  oz. ;  11  T.  4  cwt.  16  Ib.  12  oz.? 

(/)  14  sq.  yd.  8  sq.  ft.  19  sq.  in. ;  105  sq.  yd.  16  sq.  ft.  240  sq.  in. ; 
42  sq.  yd.  28  sq.  ft.  165  sq.  in.? 

(a)  86  Ib.  3  oz.  16  pwt.  7  gr. 

(b)  25  mi.  47  rd.  1  ft.  5  in. 

(c)  18  cwt.  2  Ib.  14  oz. 

(d)  78  yr.  1  mo.  3  wk.  3  da. 

(e)  65  T.  9  cwt.  28  Ib.  13  oz. 
(/)  167  sq.  yd.  136  sq.  in. 


Ans. 


SUBTRACTION  OF  DENOMINATE  NUMBERS. 

5O.      EXAMPLE'.—  From  21  rd.  2  yd.  2  ft.  6^-  in.  take  9  rd.  4  yd. 
£  in. 
SOLUTION.—  rd.          yd.        ft.  in. 


21 
9 

2 
4 

2 
0 

10* 

11     8}    1     8|  Ans. 

EXPLANATION.  —  Since  10^  inches  cannot  be  taken  from  6| 
inches,  we  must  borrow  1  foot  or  12  inches  from  the  2  feet 
in  the  next  column  and  add  it  to  the  6|.  6-J-  +  12  =  18-J. 
18-J-  inches  —  10^  inches  =  8£  inches.  Then,  0  from  the  1 
remaining  foot  =  1  foot.  4  yards  cannot  be  taken  from  2 
yards;  therefore,  we  borrow  1  rod,  or  5-J  yards,  from  21 
rods  and  add  it  to  2.  2  +  5J  =  7-J;  7£  —  4  =  3J  yards. 
9  rods  from  20  rods  =  11  rods.  Hence,  the  remainder  is  11 
rods  3£  yards  1  foot  8£  inches.  Ans. 

To  avoid  fractions  as  much  as  possible,  we  reduce  the  •} 
yard  to  inches,  obtaining  18  inches;  this  added  to  8£  inches 
gives  26J  inches,  which  equals  2  feet  2J  inches.  Then,  2 
feet  -f-  1  foot  =  3  feet  =  1  yard,  and  3  yards  +  1  yard  =  4 


18  ARITHMETIC.  §  2 

yards.     Hence,  the  above  answer  becomes  11  rods  4  yards 
0  feet  2£  inches. 

51.      EXAMPLE.— What  is  the  difference  between  3  rd.  2yd.  2ft. 

10  in.  and  47  ft.  ? 

SOLUTION.—    47  ft.  =  2  rd.  4  yd.  2  ft. 

rd.         yd.          ft.          in. 
3  2  2  10 

242 0 

0  3£  0  10 

or  3  2  4     Ans. 

To  find  (approximately)  the  interval  of  time  between 
two  dates : 

5£.      EXAMPLE. — How  many  years.months, days, and  hours  between 
4  o'clock  r.  M.  of  June  16, 1868,  and  10  o'clock  A.  M.,  September  29,  1891? 
SOLUTION. —  yr.        mo.        da.        hr. 

1891         8  28          10 

1868         5  15          16 

23        3  12         18     Ans. 

EXPLANATION. — Counting  24  hours  in  1  day,  4  o'clock  p.  M. 
is  the  16th  hour  from  the  beginning  of  the  day,  or  midnight. 
On  September  29,  8  months  and  28  days  have  elapsed,  and 

011  June  16,  5  months  and  15  days.     After  placing  the  earlier 
date  under  the  later  date,  subtract  as  in  the  previous  prob- 
lems.    Count  30  days  as  1  month. 

53.  Rule. — Place  the  smaller  quantity  under  the  larger 
quantity ',  with  like  denominations  under  each  other.  Begin- 
ning at  the  right,  subtract  successively  the  number  in  the 
subtrahend  in  each  denomination  from  the  one  above,  and 
place  the  differences  underneath.  If  the  number  in  the 
minuend  of  any  denomination  is  less  than  the  number  under 
it  in  the  subtrahend,  one  must  be  borrowed  from  the  minuend 
of  the  next  higher  denomination,  reduced,  and  added  to  it. 


EXAMPLES  EOll  PRACTICE. 
54.      From: 

(a)  125  Ib.  8  oz.  14  pwt.  18  gr.  take  96  Ib.  9  oz.  10  pwt.  4  gr. 

(b)  126  hhd.  27  gal.  take  104  hhd.  14  gal.  1  qt.  1  pt. 

(c)  65  T.  14  cwt.  64  Ib.  10  oz.  take  16  T.  11  cwt.  14  oz. 

(d)  148  sq.  yd.  16  sq.  ft.  142  sq.  in.  take  132  sq.  yd.  136  sq.  in. 


§  2  ARITHMETIC.  19 

(e)   100  bu.  take  28  bu.  2  pk.  5  qt.  1  pt. 

(/)  14  mi.  34  rd.  16  yd.  13  ft.  11  in.  take  3  mi.  27  rd.  11  yd.  4  ft.  10  in. 

'  (a)  28  Ib.  11  oz.  4  pwt.  14  gr. 

(b)  22  hhd.  12  gal.  2  qt.  1  pt. 

.    ,  (c)  49  T.  3  cwt.  63  Ib.  12  oz. 

(fi)  16  sq.  yd.  16  sq.  ft.  6  sq.  in. 

(*)  71  bu.  1  pk,  2  qt.  1  pt. 

(/)  11  mi.  7  rd.  5  yd.  9  ft.  1  in. 


MUI/TIPLICATION  OF  DENOMINATE  NUMBERS. 

55.  EXAMPLE.— Multiply  7  Ib.  5  oz.  13  pwt.  15  gr.  by  12. 
SOLUTION. —  Ib.          oz.        pwt.        gr. 

7  5  13  15 

12 

89  8~         3  12    Ans. 

EXPLANATION.—  15  grains  X  12  =  180  grains.  180^24 
=  7  pennyweights  and  12  grains  remaining.  Place  the  12 
in  the  grain  column  and  carry  the  7  pennyweights  to  the 
next.  Now,  13x12  +  7  =  163  pennyweights;  163-^-20  =  8 
ounces  and  3  pennyweights  remaining.  Then,  5x12  +  8 
=  '68  ounces ;  68-^12  =  5  pounds  and  8  ounces  remaining. 
Then,  7x12  +  5  =  89  pounds.  The  entire  product  is  89 
pounds  8  ounces  3  pennyweights  12  grains.  Ans. 

56.  Rule. — Multiply    the    number     representing    each 
denomination  by  the  multiplier  and  reduce  each  product  to 
the  next  higher  denomination,  writing"  the  remainders  under 
each  denomination,  and  carry  the  quotient  to  the  next,  as  in 
Addition  of  Denominate  Numbers. 

57.  In  multiplication  and  division  of  denominate  num- 
bers, it  is  sometimes  easier  to  reduce  the  number  to  the 
lowest  denomination  given  before  multiplying  or  dividing, 
especially  if  the  multiplier  or  divisor  is  a  decimal.     Thus, 
in  the  example  of  Art.  55,  had  the  multiplier  been  1.2,  the 
easiest  way  to  multiply  would  have  been  to  reduce  the  num- 
ber to  grains;  then,  multiply  by  1.2,  and  reduce  the  product 
to  higher  denominations.     For  example,  7  Ib.  5  oz.  13  pwt. 
15  gr.   =  43,047  gr.     43,047x1.2  =  51,656.4  gr.  =  8  Ib.  11 
oz.  12  pwt.  8.4  gr.     Also,  43,047x12  =  516,564  gr.  =  89  Ib. 
8  oz.  3  pwt.  12  gr. ,  as  above.     Either  method  may  be  used. 


20  ARITHMETIC.  §  2 

EXAMPLES  FOR  PRACTICE. 

58.      Multiply: 

(a)  15  cwt.  90  Ib.  by  5;  (b)  12  yr.  10  mo.  4  wk.  3  da.  by  14;  (c)  11  mi. 
145  rd.  by  20;  (d)  12  gal.  4  pt.  by  9;  (*)  8  cd.  76  cu.  ft.  by  15;  (/) 
4  hhd.  3  gal.  1  qt  1  pt  by  12. 

(a)  79  cwt.  50  Ib. 

(b)  180  yr.  11  mo.  2  wk. 

(c)  229  mi.  20  rd. 

(d)  112  gal.  2  qt. 

(e)  128  cd.  116  cu.  ft. 
(/)    48  hhd.  40  gal.  2  qt. 


Ans. 


DIVISION  OF  DENOMINATE  NUMBERS. 

59.      EXAMPLE.— Divide  48  Ib.  11  oz.  6  pwt.  by  8. 
SOLUTION. —  Ib.  oz.        pwt.        gr. 

8)48  11  6  0 

6  Ib.        1  oz.      8  pwt.    6  gr.     Ans. 

EXPLANATION. — After  placing-  the  quantities  as  above, 
proceed  as  follows :  8  is  contained  in  48  six  times  without  a 
remainder.  8  is  contained  in  11  ounces  once,  with  3  ounces 
remaining.  3x20  =  60;  60  +  6  =  66  pennyweights;  66 
pennyweights  -r-  8  =  8  pennyweights  and  2  remaining ; 
2x24  grains  =  48  grains;  48  grains -i- 8  =  6  grains. 
Therefore,  the  entire  quotient  is  6  pounds  1  ounce  8  pen- 
nyweights 6  grains.  Ans. 

EXAMPLE. — A  silversmith  melted  up  2  Ib.  8oz.  10  pwt.  of  silver,  which 
he  made  into  6  spoons ;  what  was  the  weight  of  each  spoon  ? 
SOLUTION. —  Ib.          oz.          pwt. 

6)2  8  10 

5  oz.         8  pwt.     8  gr.     Ans. 

EXPLANATION. — Since  we  cannot  divide  2  pounds  by  6,  we 
reduce  it  to  ounces.  2  pounds  =  24  ounces,  and  24  ounces 
-j-  8  ounces  =  32  ounces ;  32  ounces  -^6  =  5  ounces  and  2 
ounces  over.  2  ounces  =  40  pennyweights ;  40  pennyweights 
+  10  pennyweights  =  50  pennyweights,  and  50  penny- 
weights -r-  6  =  8  pennyweights  and  2  pennyweights  over.  2 
pennyweights  =  48  grains,  and  48  grains  -4-6  =  8  grains. 
Hence,  each  spoon  contains  5  ounces  8  pennyweights  8 
grains.  Ans. 


2  ARITHMETIC.  21 

6O.      EXAMPLE.— Divide  820  rd.  4  yd.  2  ft.  by  112. 

rd.    yd.  ft.    rd.  yd.  ft.     in. 

SOLUTION.—       112)820    4    2(    7    1    2    5.148  Ans. 
784 

3  6  rd.  rem. 
5.5 
180 
180 


1  9  8.0  yd. 

4 

112)  202  yd.  (  1  yd. 
112 
9  0  yd.  rem. 


2  7  0  ft. 
2ft. 

1  1  2  )  2  7  2  ft.  (  2  ft. 
224 

4  8  ft.  rem. 
12 
96 
48 

1  1  2  )  5  7  6  in.  (  5.1  4  2  8+  in.  or  5.143  in. 
560 


160 
112 
480 
448 
320 
224 
960 
896 
64 

EXPLANATION. — The  first  quotient  is  7  rods  with  36  rods 
remaining.  5.5x36  =  198 yards;  198 yards +  4 yards  =  202 
yards;  202  yards -5- 112  =  1  yard  and  90  yards  remaining. 
90X3  =  270 feet;  270  feet  +  2  feet  =  272 feet;  272 feet -^  112 
=  2  feet,  and  48  feet  remaining;  48x12  =  576  inches;  576 
inches -7- 112  =  5.143  inches,  nearly.  Ans. 

The  preceding  example  is  solved  by  long  division,  because 

1-6 


22 


ARITHMETIC. 


the  numbers  are  too  large  to  deal  with  mentally.  Instead 
of  expressing  the  last  result  as  a  decimal,  it  might  have 
been  expressed  as  a  common  fraction.  Thus,  576-^112 
—  ^TT62"  =  <H  inches.  The  chief  advantage  of  using  a  com- 
mon fraction  is  that  if  the  quotient  be  multiplied  by  the 
divisor,  the  result  will  always  be  the  same  as  the  original 
dividend. 

61  .  Rule.  —  Find  how  many  times  the  divisor  is  contained 
in  the  first  or  highest  denomination  of  the  dividend.  Reduce 
the  remainder  (if  any)  to  the  next  lower  denomination,  and 
add  to  it  the  number  in  the  given  dividend  expressing  that 
denomination.  Divide  this  new  dividend  by  the  divisor. 
The  quotient  will  be  the  next  denomination  in  the  quotient 
required.  Continue  in  this  manner  until  tJie  lowest  denomi- 
nation is  readied.  The  successive  quotients  will  constitute 
the  entire  quotient. 

EXAMPLES  FOR  PRACTICE. 
62.      Divide: 

(a)  376  mi.  276  rd.  by  22;  (l>)  1,137  bu.  3  pk.  4  qt.  1  pt.  by  10; 
(c)  84  cwt.  48  Ib.  49  oz.  by  16;  (d)  78  sq.  yd.  18  sq.  ft.  41  sq.  in.  by  18; 
(*)  148  mi.  64  rd.  24  yd.  by  12;  (/)  100  T.  16  cwt.  18  Ib.  11  oz.  by  15; 
(g)  36  Ib.  18  oz.  18  pwt.  14  gr.  by  8;  (h)  112  mi.  48  rd.  by  100. 


Ans.  < 


(a) 
(*) 

M 

(</) 
« 

(/) 


17  mi.  41T7T  rd. 

113  bu.  3  pk.  1  qt.  |  pt. 

5  cwt.  28  Ib.  3T^  oz. 

4  sq.  yd.  4  sq.  ft.  2T5F  sq.  in. 
12  mi.  112  rd.  2  yd. 

6  T.  14  cwt.  41  Ib.  3{|  oz. 
4  Ib.  8  oz.  7  pwt.  7f  gr. 
lmi..38|f  rd. 


63.  Involution  is  the  process  of  rrmltiplying  a  number 
by  itself  one  or  more  times.  The  product  obtained  by 
multiplying  a  number  by  itself  is  called  a  power  of  that 
number. 

Thus,  the  second  power  of  3  is  9,  since  3x3  are  9. 


g  2  ARITHMETIC.  23 

The  third  power  of  3  is  27,  since  3x3x3  are  27. 
The  fifth  power  of  2  is  32,  since  2  X  2  X  2  X  2  X  2  are  32. 

64.  An  exponent  is  a  small  figure  placed  to  the  right 
and  a  little  above  a  number  to  show  to  what  power  it  is  to  be 
raised,  or  how  many  times  the  number  is  to  be  used  as  a 
factor,  as  the  small  figures  2>  3»  and  5  below. 

Thus,  32  =  3X3  =  9. 

33  =  3  X  3  X  3  =  27. 

25  =  2X2X2X2X2  =  32. 

65.  The  root  of  a  mimber  is  that  number  which,  used 
the  required  number  of  times  as  a  factor,   produces  the 
number.     In  the  above  cases  3  is  a  root  of  9,  since  3x3  are  9. 
It  is  also  a  root  of  27,  since  3x3x3  are  27.     Also,  2  is  a 
root  of  32,  since  2  X  2  X  2  X  2  X  2  are  32. 

66.  The    second   power    of    a    number    is    called    its 
square. 

Thus,  52  is  called  the  square  of  5,  or  5  squared,  and  its 
value  is  5  X  5  =  25. 

67.  The  third  power  of  a  number  is  called  its  cube. 
Thus,  53  is  called  the  cube  of  5,  or  5  cubed,  and  its  value 

is  5x5x5  =  125. 

To  find  any  power  of  a  number  : 

68.  EXAMPLE.  —  What  is  the  third  power,  or  cube,  of  35  ? 

SOLUTION.—    -.  35  X  35  X  35, 

or  3  5 

35 


105 


1225 
35 

6125 
3675 


cube  =  42875    Ans. 


24 


ARITHMETIC. 


EXAMPLE. — What  is  the  fourth  power  of  15  ? 
SOLUTION.—  15  X  15  X  15  X  15, 

or  15 


75 
15 
225 
1  5 

1125 
225 
3375 

lj> 

16875 
3375 

fourth  power  =  50625    Ans. 
69.      EXAMPLE.—    1.2s  =  what? 
SOLUTION.—  1.2x1.2x1-2, 

or  1.2 

1.2 
1.44 
1.2 
288 
144 


cube  -  1.7  2  8    Ans. 


7O.      EXAMPLE. — What  is  the  third  power,  or  cube,  of  £  ? 


SOLUTION.—      (f)8  =  f  X  I  X I  = 


3X3X3 


=  irrV 


8X8X8 

71.  Rule. — I.  71?  raise  a  whole  number  or  a  decimal  to 
any  power,  use  it  as  a  factor  as  many  times  as  there  are  units 
in  the  exponent. 

II.  To  raise  a  fraction  to  any  power,  raise  both  the  numer- 
ator and  denominator  to  the  power  indicated  by  the  exponent. 


EXAMPLES  FOR  PRACTICE. 

Raise  the  following  to  the  powers  indicated : 


(a) 

(*) 
(*) 

(</) 

(') 

CO 

(f) 
(*) 


852. 

(H)2- 

6.52. 

14*. 

(t)3. 

(I)3- 

(I)3- 

1.45. 


Ans. 


(a)      7,225. 


M 

(</) 

M 


42.25. 
38,416. 

H- 
iff- 

»|3. 

5.37824. 


§  2  ARITHMETIC.  25 

EVOLUTION. 

73.  Evolution  is  the  reverse  of  involution.     It  is  the 
process  of  finding"  the  root  of  a  number  which  is  considered  as 
a  power. 

74.  The  square  root  of  a  nu'mber  is  that  number  which, 
when  used  twice  as  a  factor,  prodtices  the  number. 

Thus,  2  is  the  square  root  of  4,  since  2x2,  or  22  =  4. 

75.  The  cube  root  of  a  number  is  that  number  which, 
when  used  three  times  as  a  factor,  produces  the  number. 

Thus,  3  is  the  cube  root  of  27,  since  3x3x3,  or  33  =  27. 

76.  The  radical  sign  y,  when  placed  before  a  number, 
indicates  that  some  root  of  that  number  is  to  be  found. 

77.  The  index  of  the  root  is  a  small  figure  placed  over 
and  to  the  left  of  the  radical  sign,  to  show  what  root  is  to  be 
found. 

Thus,   ^100  denotes  the  square  root  of  100. 
Vl25  denotes  the  cube  root  of  125. 
y  256  denotes  \h.e  fourth  root  of  256,  and  so  on. 

78.  When  the  square  root  is  to  be  extracted,  the  index  is 
generally  omitted.     Thus,  l/lOO  indicates  the  square  root  of 
100.     Also,    4/225  indicates  the  square  root  of  225. 


SQUARE    ROOT. 

79.  The  largest  number  that  can  be  written  with  one 
figure  is  9,  and  92  =  81;  the  largest  number  that  can  be 
written  with  two  figures  is  99,  and  992  =  9,801;  with  three 
figures  999,  and  9992  =  998,001 ;  with/<??/r  figures  9,999,  and 
9,9992  =  99,980,001,  etc. 

In  each  of  the  above  it  will  be  noticed  that  the  square 
of  the  number  contains  just  twice  as  many  figures  as  the 
number.  , 

In  order  to  find  tne  square  root  of  a  number,  the  first 
step  is  to  find  how  many  figures  there  will  be  in  the  root. 


26  ARITHMETIC.  §  2 

This  is  done  by  pointing1  off  the  number  into  periods  of  two 
figures  each,  beginning  at  the  rig/it.  The  number  of  periods 
will  indicate  the  number  of  figures  in  the  root. 

Thus,  the  square  root  of  83,740,801  must  contain  4  figures, 
since,  pointing  off  the  periods,  we  get  83'74'08'01,  or  4 
periods ;  consequently,  there  must  be  4  figures  in  the  root. 
In  like  manner,  the  square  root  of  50,625  must  contain  3  fig- 
ures, since  there  are  (5'06'25)  3  periods. 

8O.      EXAMPLE.— Find  the  square  root  of  31,505,769. 

roof 
SOLUTION.—        (a)         5  3  1'5  0'5  7'6  9  (  5  6  1  3     Aiis. 

_5        (l>)  2_5 
(it)  100       (c)   650 
6  636 

1  0  6       (,•)     1457 
6  1121 


1120  88669 
1  38669 

1121  0 
1 

11220 
3 


11223 

EXPLANATION. — Pointing-  off  into  periods  of  two  figures 
each,  it  is  seen  that  there  are  four  figures  in  the  root.  Now, 
find  the  largest  single  number  whose  square  is  less  than  or 
equal  to  31,  the  first  period.  This  is  evidently  5,  since  62  =  36, 
which  is  greater  than  31.  Write  it  to  the  right,  as  in  long 
division,  and  also  to  the  left  as  shown  at  (a).  This  is  the 
first  figure  of  the  root.  Now,  multiply  the  5  at  (a)  by  the  5 
in  the  root,  and  write  the  result  under  the  first  period,  as 
shown  at  (#).  Subtract,  and  obtain  6  as  a  remainder. 

Bring  down  the  next  period,  50,  and  annex  it  to  the 
remainder,  6,  as  shown  at  (c],  which  we  call  the  dividend. 
Add  the  root  already  found  to  the  5  at  (a),  getting  10,  and 
annex  a  cipher  to  this  10,  thus  making  it  100,  which  we  call 
the  -trial  divisor.  Divide  the  dividend  (c)  by  the  trial 
divisor  (</)  and  obtain  6,  which  is  probably  the  next  figure 
of  the  root.  Write  6  in  the  root,  as  shown,  and  also  add  it 


§  2  ARITHMETIC.  27 

to    100,  the  trial   divisor,    making  it    106.     This    is    called 
the  complete  divisor. 

Multiply  this  by  6,  the  second  figure  in  the  root,' and  sub- 
tract the  result  from  the  dividend  (c).  The  remainder  is  14, 
to  which  annex  the  next  period,  making  it  1,457,  as  shown 
at  (e),  which  we  call  the  new  dividend.  Add  the  second 
figure  of  the  root  to  the  complete  divisor,  100,  and  annex  a 
cipher,  thus  getting  1,120.  Dividing  1,457  by  1,120,  we  get 
1  as  the  next  figure  of  the  root.  Adding  this  last  figure 
of  the  root  to  1,120,  multiplying  the  result  by  it,  and  sub- 
tracting from  1,457,  the  remainder  is  336. 

Annexing  the  next  and  last  period,  69,  the  result  is  33,669. 
Now,  adding  the  last  figure  of  the  root  to  1,121,  and  annex- 
ing a  cipher  as  before,  the  result  is  11,220.  Dividing  33,669 
by  11,220,  the  result  is  3,  the  fourth  figure  in  the  root. 
Adding  it  to  11,220  and  multiplying  the  sum  by  it,  the  result 
.is  33,669.  Subtracting,  there  is  no  remainder;  hence, 
4/31,505,769  =  5,613. 

81.  The  square  of  any  number  wholly  decimal  always 
contains  twice  as  many  figures  as  the   number   squared. 
For  example,  .I2  =  .01,   .132  =  .0169,   .7512  =  .564001,  etc. 

82.  It  will  also  be  noticed  that  the  number  squared  is 
always  less  than  the  decimal.     Hence,  if  it  be  required  to 
find  the  square  root  of  a  decimal,  and  the  decimal  has  not  an 
even  number  of  figures  in  it,  annex  a  cipher.     The  best  way 
to  determine  the  number  of  figures  in  the  root  of  a  decimal 
is  to  begin  at  the  decimal  point,  and,   going  towards  the 
right,  point  off  the  decimal  into  periods  of  two  figures  each. 
Then,  if  the  last  period  contains  but  one  figure,  annex  a 
cipher. 

83.  EXAMPLE.— What  is  the  square  root  of  .000576  ? 

root 
SOLUTION.—  2  .0  O'O  5'7  6  ( .0  2  4    Ans. 

2I_  4 

40  176 

_4  176 

44  0 


28  ARITHMETIC.  §  2 

EXPLANATION. — Beginning  at  the  decimal  point,  and  point- 
ing off  the  number  into  periods  of  two  figures  each,  it  is  seen 
that  the  first  period  is  composed  of  ciphers ;  hence,  the  first 
figure  of  the  root  must  be  a  cipher.  The  remaining  portion 
of  the  solution  should  be  perfectly  clear  from  what  has 
preceded. 

84.  If  the  number  is  not  a  perfect  power,  the  root  will 
consist  of  an  interminable  number  of  decimal  places.     The 
result  may  be  carried  to  any  required  number  of  decimal 
places  by  annexing  periods   of   two   ciphers   each   to   the 
number. 

85.  EXAMPLE. — What  is  the  square  root  of  3  ?    Find  the  result  to 

five  decimal  places. 

root 
3.0  O'O  O'O  O'O  O'O  0  ( 1.7  3  20  5  +    Ans. 

2~00 
189 
1100 
1029 
7100 
6924 
1760000 
1732025 
27975 


346405 

EXPLANATION. — Annexing  five  periods  of  two  ciphers  each 
to  the  right  of  the  decimal  point,  the  first  figure  of  the  root 
is  1.  To  get  the  second  figure  we  find  that,  in  dividing  200 
by  20,  it  is  10.  This  is  evidently  too  large. 

Trying  9,  we  add  9  to  20,  and  multiply  29  by  9 ;  the  result 
is  261,  a  result  which  is  considerably  larger  than  200;  hence, 
9  is  too  large.  In  the  same  way  it  is  found  that  8  is  also  too 
large.  Trying  7,  7  times  27  are  189,  a  result  smaller  than 


§  2  ARITHMETIC.  29 

200 ;  therefore,  7  is  the  second  figure  of  the  root.  The  next 
two  figures,  3  and  2,  are  easily  found.  The  fifth  figure  in 
the  root  is  a  cipher,  since  the  trial  divisor,  34,640,  is  greater 
than  the  new  dividend,  17,600.  In  a  case  of  this  kind  we 
annex  another  cipher  to  34,640,  thereby  making  it  346,400, 
and  bring  down  the  next  period,  making  the  17,600, 
1,760,000.  The  next  figure  of  the  root  is  5,  and,  as  we  now 
have  five  decimal  places,  we  will  stop. 

The  square  root  of   3   to   five   decimal  places   is,   then, 
1. 73205 +  . 

86.      EXAMPLE. — What  is  the  square  root  of  .3  to  five  decimal 

places  ? 

root 

SOLUTION.—     5  .3  O'O  O'O  O'O  O'O  0  (  .54772+    Ans. 

__5  ^5 

foo       Too 

4  416 

104  8400 

4  7609 

1080  79100 

7  76629 

1087  247100 

7  219084 

10940  28016 
7 


10947 

7 

109540 

2 

109542 

EXPLANATION. — In  the  above  example  we  annex  a  cipher 
to  .3,  making  the  first  period  .30,  since  every  period  of  a 
decimal,  as  was  mentioned  before,  must  have  two  figures 
in  it.  The  remainder  of  the  work  should  be  perfectly 
clear. 

87.  If  it  is  required  to  find  the  square  root  of  a  mixed 
number,  begin  at  the  decimal  point,  and  point  off  the  periods 
both  ways.  The  manner  of  finding  the  root  will  then  be 
exactly  the  same  as  in  the  previous  cases. 


30  ARITHMETIC. 

88.      EXAMPLE. — What  is,  the  square  root  of  258.2449? 

root 
SOLUTION.— 


1 

2'5  8.2  4'4  9 

1 

1 

To 

158 

6 

1  56 

26 

22449 

6 

22449 

3200 

0 

7 

3207 

EXPLANATION. — In  the  above  example,  since  320  is  greater 
than  224,  we  place  a  cipher  for  the  third  figure  of  the  root, 
and  annex  a  cipher  to  320,  making  it  3,200.  Then,  bringing 
down  the  next  period,  49,  7  is  found  to  be  the  fourth  figure 
of  the  root.  Since  there  is  no  remainder,  the  square  root  of 
258.2449  is  16.07. 

89.  Proof. —  To  prove  square  root,    square   the    result 
obtained.     If  the  number  is  an  exact  power,   the  square  of 
the  root  ivill  equal  it;  if  it  is  not  an  exact  poiver,  the  square 
of  the  root  will  very  nearly  equal  it. 

90.  Rule. — I.     Begin  at  units  place,  and  separate  the 
number  into  periods  of  two  figures  each,  proceeding  from 
left  to  right  with  the  decimal  part,  if  there  be  any. 

II.  Find  the  greatest  number  whose  square  is  contained 
in  the  first,  or  left-hand,  period.      Write  this  number  as  the 
first  figure  in  the  root;  also,  write  it  at  the  left  of  the  given 
number. 

Multiply  this  number  at  the  left  by  the  first  figure  of  the 
root,  and  subtract  the  result  from  the  first  period;  then, 
annex  the  second  period  to  the  remainder. 

III.  Add  the  first  figure  of  the  root  to  the  number  in  the 
first  column  on  the  left,  and  annex  a  cipher  to  the  result; 
this  is  the  trial  divisor.     Divide  the  dividend  by  the  trial 
divisor  for  the  second  figure  in  the  root,  and  add  this  figure 
to  the  trial  divisor  to  form  the  complete  divisor.     Multiply 
the  complete  divisor  by  the  second  figure  in  the  root,  and 


§  2  ARITHMETIC.  31 

subtract  this  result  from  the  dividend.  (If  this  result  is  larger 
than  the  dividend,  a  smaller  number  must  be  tried  for  the 
seeond figure  of  the  root.)  Now  bring  doivn  tJie  tJdrd  period, 
and  annex  it  to  the  last  remainder  for  a  new  dividend.  Add 
tlic  second  figure  of  the  root  to  the  complete  divisor ',  and  annex 
a  cipher  for  a  new  trial  divisor. 

IV.  Continue  in  this  manner  to  the  last  period,   after 
wJiich,    if  any  additional  places  in  the  root  are  required, 
bring  down  cipher  periods,  and  continue  the  operation. 

V.  If  at  any  time  the  trial  divisor  is  not  contained  in  the 
dividend,  place  a  cipher  in  the  root,  annex  a  cipher  to  the 
trial  divisor,  and  bring  down  another  period. 

VI.  If  the  root  contains  an  interminable  decimal,  and  it 
is  desired  to  terminate  the  operation  at  some  point,  say,  the 
fourth  decimal  place,  carry  the  operation  one  place  further, 
and  if  tJie  fifth  figure  is  5  or  greater,  increase  the  fourth 
figure  by  1  and  omit  the  sign  -j- . 

91.  Short  Method.— If  the  number  whose  root  is  to  be 
extracted  is  not  an  exact  square,  the  root  will  be  an  inter- 
minable decimal.  It  is  then  usual  to  extract  the  root  to  a 
certain  number  of  decimal  places.  In  such  cases,  the  work 
may  be  greatly  shortened  as  follows:  Determine  to  how 
many  decimal  places  the  work  is  to  be  carried,  say  5,  for 
example ;  add  to  this  the  number  of  places  in  the  integral 
part  of  the  root,  say  2,  for  example,  thus  determining  the 
number  of  figures  in  the  root,  in  this  case  5  +  2  =  7.  Divide 
this  number  by  2  and  take  the  next  higher  number.  In  the 
above  case,  we  have  7^2  =  3-J ;  hence,  we  take  4,  the  next 
higher  number.  Now  extract  the  root  in  the  usual  manner 
until  the  same  number  of  figures  has  been  obtained  as  was 
expressed  by  the  number  obtained  above,  in  this  case  4. 
Then  form  the  trial  divisor  in  the  usual  manner,  but  omit- 
ting to  add  the  cipher;  divide  the  last  remainder  by  the 
trial  divisor  as  in  long  division,  obtaining  as  many  figures 
of  the  quotient  as  there  are  remaining  figures  of  the  root,  in 
this  case  7  —  4  =  3.  The  remainder  so  obtained  is  the 
remaining  figures  of  the  root. 


32  ARITHMETIC.  §  2 

Consider  the  example  in  Art.  86.  Here  there  are  5  fig- 
ures in  the  root.  We  therefore  extract  the  root  to  3  places 
in  the  usual  manner,  obtaining  .  547  for  the  first  three  root 
figures.  The  next  trial  divisor  is  1,094  (with  the  cipher 
omitted)  and  the  last  remainder  is  791.  Then,  791-4-1,094 
=  .123,  and  the  next  two  figures  of  the  root  are  72,  the 
whole  root  being  .  54772  -j-  .  Always  carry  the  division  one 
place  further  than  desired,  and  if  the  last  figure  is  5  or 
greater  increase  the  preceding'  figure  by  1.  This  method 
should  not  be  used  unless  the  root  contains  five  or  more 
figures. 

If  the  last  figure  of  the  root  found  in  the  regular  man- 
ner is  a  cipher,  carry  the  process  one  place  further  before 
dividing  as  described  above. 


(a) 


EXAMPLES   FOB  PRACTICE. 

Find  the  square  root  of: 

186,624. 

(b)  2,050,624. 

(c)  29,855,296. 
(;/)    .0116964. 
(e)      198.1369. 

(/)  994,009.  Ans. 

(«£")  2.375  to  four  decimal  places. 

(h)  1.625  to  three  decimal  places. 

(/)  .3025. 

(/)  .571428. 

(k)  .78125. 


(a) 

432. 
1,432. 

M 

5,464. 
.1081  + 

M 
(/) 

14.0761. 
997. 
1.5411. 

(1 
(0 
GO 
W 

1.275. 
.55. 

.7559  + 
.8839. 

CUBE  ROOT. 

93.  In  the  same  manner  as  in  the  case  of  square  root, 
it  can  be  shown  that  the  periods  into  which  a  number 
is  divided,  whose  cube  root  is  to  be  extracted,  must  con- 
tain three  figures,  except  that  the  first,  or  left-hand,  period 
of  a  whole  or  mixed  number  may  contain  one,  two,  or  three 
figures. 


ARITHMETIC. 


33 


94.        EXAMI'I 

SOLUTION.  — 
(1) 

7 
7 

.F.  —  What  is  the 

(2) 

49 

98 

cube  root  of  375,741,853,696  ? 

(3)                         root 
375'741'853'696(7216 
343 

14 

7  ' 

14700 
424 
15124 

428 

32741 
30248 

210 
2 
212 
2 
214 
2 

2493853 
1557361 

1555200 
2161 

936492696 
936492696 

1557361 
2162 

0 

2160 
1 

155952  30  0 
129816 

2161 
1 

156082116 

Ans. 


2162 
1 

21630 
6 

21636 

EXPLANATION. — Write  the  work  in  three  columns,  as  fol- 
lows :  On  the  right,  place  the  number  whose  cube  root  is  to 
be  extracted,  and  point  it  off  into  periods  of  three  figures 
each.  Call  this  column  (3).  Find  the  largest  number  whose 
cube  is  less  than  or  equal  to  the  first  period,  in  this  case  7. 
Write  the  7  on  the  right  as  shown,  for  the  first  figure  of  the 
root,  and  also  on  the  extreme  left  at  the  head  of  column  (1). 
Multiply  the  7  in  column  (1)  by  the  first  figure  of  the  root, 
7,  and  write  the  product,  49,  at  the  head  of  column  (2).  Mul- 
tiply the  number  in  column  (2)  by  the  first  figure  of  the 
root,  7,  and  write  the  product,  343,  under  the  figures  in  the 
first  period.  Subtract  and  bring  down  the  next  period, 
obtaining  32,741  for  the  dividend.  Add  the  first  figure 
of  the  root  to  the  number  in  column  (1),  obtaining  14, 
which  call  the  first  correction.  Multiply  the  first  correction 
by  the  first  figure  of  the  root,  add  the  product  to  the 
number  in  column  (2),  and  obtain  147.  Add  the  first  figure 
of  the  root  to  the  first  correction,  and  obtain  21,  which  call 


34  ARITHMETIC.  §  2 

the  second  correction.  Annex  two  ciphers  to  the  number 
in  column  (2),  and  obtain  14,700  for  the  trial  divisor;  also 
annex  one  cipher  to  the  second  correction,  and  obtain  210. 

32  741 
Dividing  the  dividend  by  the  trial  divisor,  we  obtain  T- 


)  and  write  the  2  as  the  second  figure  of  the  root. 
Add  the  2  to  the  second  correction,  and  obtain  212,  which 
multiplied  by  the  second  figure  of  the  root  and  added  to 
the  trial  divisor,  gives  15,124,  the  complete  divisor.  This 
last  result  multiplied  by  the  second  figure  of  the  root  and 
subtracted  from  the  dividend,  gives  a  remainder  of  2,493. 
Annexing  the  third  period,  we  obtain  2,493,853  for  the  new 
dividend.  Adding  the  second  figure  of  the  root  to  the 
number  in  column  (1)  we  get  214  as  the  new  first  correction  ; 
this  multiplied  by  the  second  figure  of  the  root  and  added 
to  the  complete  divisor,  gives  15,552.  Adding  the  second 
figure  of  the  root  to  the  first  new  correction  gives  216  as  the 
second  new  correction.  Annexing  two  ciphers  to  the  num- 
ber in  column  (2)  gives  1,555,200,  the  new  trial  divisor. 
Annexing  one  cipher  to  the  second  new  correction  gives 
2,100.  Dividing  the  new  dividend  by  the  new  trial  divisor 

.    .      2,493,853 
we  obtain  ^  =  1  +  ,  and  write  1  as  the  third  figure 


of  the  root.     The  remainder  of  the  work  should  be  perfectly 
clear  from  what  has  preceded. 

95,  In  extracting  the  cube  root  of  a  decimal,  proceed  as 
above,  taking  care  that  each  period  contains  three  figures. 
Begin  the  pointing  off  at  the  decimal  point,  going  towards 
the  right.      If  the  last  period  does  not  contain  three  figures, 
annex  ciphers  until  it  does. 

96.  EXAMPLE.—  What  is  the  cube  root  of  .009129329  ? 

root 

SOLUTION.—  2  4  .0  0  9'1  2  9'3  2  9  (  .2  0  9 

2_  __8  8 

4~  120000  1  129329 

2_  5481  1  129329 

600  125481  0 

9 
609 


§  2  ARITHMETIC.  35 

EXPLANATION. — Beginning  at  the  decimal,  and  pointing 
off  as  shown,  the  largest  number  whose  cube  is  .less  than  9 
is'  seen  to  be  2;  hence,  2  is  the  first  figure  of  the  root. 
When  finding  the  second  figure,  it  is  seen  that  the  trial 
divisor,  1,200,  is  greater  than  the  dividend;  hence,  write  a 
cipher  for  the  second  figure  of  the  root ;  bring  down  the 
next  period  to  form  the  new  dividend ;  annex  two  ciphers 
to  the  trial  divisor  to  form  a  new  trial  divisor ;  also,  annex 
one  cipher  to  the  60  in  the  first  column.  Dividing  the  new 
dividend  by  the  new  trial  divisor,  we  get  -VVWuV"  =  9  +  > 
and  write  9  as  the  third  figure  of  the  root.  Complete  the 
work  as  before. 

9 7 .      EXAMPLE.— What  is  the  cube  root  of  78,292. 892952  ? 

SOLUTION. — 

root 

4  16  78'292.892'952(42.78 

4  32  64 

8         4800          14292 
4  244          10088 

120         5044  4204892 

_2          248  3766483 

122         529200          438409952 
2  8869          438409952 


124         538069 
2  8918 


1260        54698700 
7          102544 


1267   .     54801244 

7 


1274 

7 


EXPLANATION. — Since  the  above  is  a  mixed  number,  begin 
at  the  decimal  point  and  point  off  periods  of  three  figures 


36  ARITHMETIC.  §  2 

each,  in  both  directions.  The  first  period  contains  but  two 
figures,  and  the  largest  number  whose  cube  is  less  than  78 
is  4  ;  consequently,  4  is  the  first  figure  of  the  root.  The 
remainder  of  the  work  should  be  perfectly  clear.  When 
dividing  the  dividend  by  the  trial  divisor  for  the  third  figure 
of  the  root,  the  quotient  was  8  -f-  >  but,  on  trying  it,  it  wras 
found  that  8  was  too  large,  the  complete  divisor  being  con- 
siderably larger  than  the  trial  divisor.  Therefore,  7  was 
used  instead  of  8. 

98.      EXAMPLE.  —  What  is  the  cube  root  of  5  to  five  decimal  places  ? 

SOLUTION.  — 

root 

1  1  5.0  0  O'O  0  O'O  0  O'O  0  O'O  0  0  (  1.70997+ 

1  2_  1_  - 

2  300  4000 
1  259  3913 


30  559  87000000 

7  308  78443829 

37  8670000  8556171000 

7  45981  7889992299 

44  8715981  666178701000 

7  46062  614014317973 

5TOO  876204300  52164383027 

9  461511 


5109       876665811 
9          461592 


5118       87712740300 
9  3590839 


51270      87716331139 
9 


51279 


512970 

7 

512977 


§  2  ARITHMETIC.  37 

EXPLANATION. — In  the  above  example,  we  annex  five 
periods  of  ciphers,  of  three  ciphers  each,  to  the  5  for  the 
decimal  part  of  the  root,  placing  the  decimal  point  between 
the  5  and  the  first  cipher.  Since  it  is  easy  to  see  that  the 
next  figure  of  the  root  will  be  5,  we  increase  the  last  figure 
by  1,  obtaining  1.70998  for  the  correct  root  to  5  decimal 
places.  Ans. 

99.  EXAMPLE. — What  is  the  cube  root  of  .5  to  four  decimal 
places  ? 

SOLUTION. — 

root 

7  49  ,  .500'000'000'000(.7937  + 

7  98  343 


14        14700        157000 
7         1971        150039 


210       16671  6961000 

9        2052  5638257 


219       1872300        1322743000 
9          7119'       1321748953 


228       1879419  994047 

9          7128 


2370     188654700 
3         166579 


188821279 


EXPLANATION. — In  the  above  example,  we  annex  two 
ciphers  to  the  .5  to  complete  the  first  period,  and  three 
periods  of  three  ciphers  each.  The  cube  root  of  500  is  7 ; 
this  we  write  as  the  first  figure  of  the  root.  The  remainder 
of  the  work  should  be  perfectly  plain  from  the  explanations 
of  the  preceding  examples. 

1-7 


38 


ARITHMETIC. 


1OO.      EXAMPLE. — What  is  the  cube  root  of  .05  to  four  decimal 


places  ? 
SOLUTION.- 


3 
_3 

6 
J 

90 
6 

96 
6 

103 
6 
1080 


388800 

8704 
397504 

8768 

4  0  6  2  7  2.0  0 

44176 

40671376 


root 

.0  5  O'O  0  O'O  0  O'O  0  0  (.3  6  8  4  + 
27 

23000 
19656 
3344000 
3180032 
163968000 
162685504 
1282496 


101.  Proof. —  To  prove   cube   root,    cube   the  result  ob- 
tained.    If  the  given  number  is  an  exact  poiver,  the  cube  of 
the  root  will  equal  it;  if  not  an  exact  power,  the  cube  of  the 
root  will  very  nearly  equal  it. 

102.  Rule. — I.     Arrange   tJie  work   in  three  columns, 
placing  the  number,  w/iose  cube  root  is  to  be  extracted,  in  tJic 
third,   or   rigJit-liand,    column.     Begin  at   units  place,  and 
separate  the  number  into  periods  of  three  figures  each,  pro- 
ceeding from  the  decimal  point  towards  the  right  with  the 
decimal  part,  if  there  is  any. 

II.  Find  the  greatest  number  whose  cube  is  not  greater 
than  the  number  in  the  first  period.  Write  this  number  as  the 
first  figiire  of  the  root ;  also,  write  it  at  the  head  of  the  first 
column.  Multiply  the  number  in  the  first  column  by  the  first 
figure  in  the  root,  and  write  the  result  in  the  second  columti. 
Multiply  the  number  in  the  second  column  by  the  first  figure 
of  the  root ;  subtract  the  product  from  the  first  period,  and 


§  2  ARITHMETIC.  39 

annex  the  second  period  to  the  remainder  for  a  new  dividend  ; 
add  the  first  figure  of  the  root  to  the  number  in  the  first 
column  for  the  first  correction.  Multiply  the  first  correction 
by  the  first  figure  of  the  root,  and  add  the  product  to  the 
number  in  the  second  column.  Add  the  first  figure  of  the 
root  to  the  first  correction  to  form  the  second  correction. 
Annex  one  cipher  to  the  second  correction  and  two  ciphers  to 
the  last  number  in  the  second  column  ;  the  last  number  in  the 
second  column  is  the  trial  divisor. 

III.  Divide  the  dividend  by  the  trial  divisor  to  find  the 
second  figure  of  the  root.     Add  the  second  figure  of  the  root 
to  the  number  in  the  first  column,  multiply  the  sum  by  the 
second  figure  of  the  root,  and  add  the  result  to  the  trial 
divisor  to  form  the  complete  divisor.     Midtiply  the  complete 
divisor  by  the  second  figure  of  the  root,  subtract  the  result 
from  the  dividend  in  the  third  column,  and  annex  the  third 
period  to  the  remainder  for  a  new  dividend.     Add  the  second 
figure  of  the  root  to  the  number  in  the  first  column  to  form 
the  first  correction  ;  multiply  the  first  correction  by  tJie  second 
figure  of  the  root,  and  add  the  product  to  the  complete  divisor. 
Add  the  second  figtire  of  the  root  to  the  first  correction  to 
form  the  second  correction.     Annex  one  cipher  to  the  second 
correction  and  tivo  ciphers  to  the  last  number  in  the  second 
column  to  form  the  new  trial  divisor. 

IV.  If  there  are  more  periods  to  be  brought  doivn,  proceed 
as  before.     If  there  is  a  remainder  after  tJie  root  of  the  last 
period  has  been  found,  annex  ciplier  periods,  and  proceed  as 
before.    The  figures  of  the  root  thus  obtained  will  be  decimals. 

V.  If  the  root  contains  an  interminable  decimal,  and  it  is 
desired  to  terminate  the  operation  at  some  point,  say,  the 
fourth  decimal  place,  carry  the  operation  one  place  further, 
and  if  the  fifth  figure  is  5  or  greater,  increase  the  fourtJi 
figure  by  1  and  omit  the  sign  -{- . 

1O3.  Art.  91  can  be  applied  to  cube  root  (or  any 
other  root)  as  well  as  to  square  root.  Thus,  in  the  exam- 
ple, Art.  98,  there  are  to  be  5  +  1  =  6  figures  in  the 
root.  Extracting  the  root  in  the  usual  manner  to  G  -^  2  =  3, 


40  ARITHMETIC.  §  2 

say  4,  figures,  we  get  for  the  first  four  figures  1,709.  The 
last  remainder  is  8,556,171,  and  the  next  trial  divisor  with 
the  ciphers  omitted  is  8,762,043.  Hence,  the  next  two 
figures  of  the  root  are  8,556,171 -r- 8, 762, 043  =.976,  say  .98. 
Therefore,  the  root  is  1.70998. 


ROOTS  OF  FRACTIONS. 

104.  If  the  given  number  is  in  the  form  of  a  fraction, 
and  it  is  required  to  find  some  root  of  it,  the  simplest  and 
most  exact  method  is  to  reduce  the  fraction  to  a  decimal  and 
extract  the  required  root  of  the  decimal.     If,  however,  the 
numerator   and   denominator   of   the   fraction   are   perfect 
powers,  extract  the  required  root  of  each  separately,  and 
write  the  root  of  the  numerator  for  a  new  numerator,  and 
the  root  of  the  denominator  for  a  new  denominator. 

105.  EXAMPLE.  —  What  is  the  square  root  of  -fa  ? 

/  n  A/  q 

SOLUTION.—  r   64  =     /^  =  *'     Ans' 

106.  EXAMPLE.  —  What  is  the  square  root  of  f  ? 
SOLUTION.—  Since  f  =.625,   |/f  =    |/.625  =.7906.     Ans. 

107.  EXAMPLE.—  What  is  the  cube  root  of  f|  ? 
SOLUTION.-  =  -         =  *.     Ans. 


108.  EXAMPLE.  —  What  is  the  cube  root  of  £  ? 
SOLUTION.—  Since  £  =.25,  ^  -    ^725  =.62996  +  .     Ans. 

109.  Rule.  —  Extract  the  required  root  of  the  numer- 
ator and  denominator  separately  ;  or,  reduce  the  fraction  to 
a  decimal,  and  extract  the  root  of  the  decimal. 


EXAMPLES  FOR  PRACTICE. 

HO.      Find  the  cube  root  of : 

(a)   f 


(b)  2  to  five  decimal  places. 

(f)  4, 180, 769, 192. 462  to  five  decimal  places.    A^ 

0  r* 

(/)  513,229.783302144  to  three  decimal  places. 


(b)  1. 25992  +  . 

(c)  1,610.96238. 

(d)  .8862  +  . 

(e)  .7211  +  . 
(/)  80.064. 


§  2  ARITHMETIC.  41 

TO  EXTRACT    OTHER    ROOTS    THAN    THE  SQUARE 
AND  CUBE  ROOTS. 

111.      EXAMPLE.—  What  is  the  fourth  root  of  256  ? 
SOLUTION.—  4/256  =  16. 

4/16  =  4. 
Therefore,  /f/256  =  4.     Ans. 


In  this  example,  f56,  the  index  is  4,  which  eqiials  2x2. 
The  root  indicated  by  2  is  the  square  root  ;  therefore,  the 
square  root  is  extracted  twice. 


EXAMPLE.—  What  is  the  sixth  root  of  64  ? 
SOLUTION.—  4/64  =  8. 

^8  =  2. 
Therefore,  -j/64  =  2.     Ans. 

In  this  example,  {/(M,  the  index  is  G,  which  equals  2  X  3. 
The  root  indicated  by  3  is  the  cube  root;  therefore,  the 
sqtiare  and  cube  roots  are  extracted  in  succession. 

113.  Rule.  —  Separate  the  index  of  the  required  root  into 
its  factors  (2's  and  3's),  and  extract,  successively,  the  roots 
indicated  by  the  several  factors  obtained.      The  final  result 
zvill  be  the  required  root. 

114.  EXAMPLE.—  What  is  the  sixth  root  of  92,873,580  to    two 
decimal  places  ? 

SOLUTION.  —  6  =  3x2.  Hence,  extract  the  cube  root,  and  then 
extract  the  square  root  of  the  result,  f  92,  873,  580  =  452.8601,  and 
V45278601  =  21.28-H.  Ans. 

115.  It  matters  not  which  root  is  extracted  first,  but  it 
is  probably  easier  and  more  exact  to  extract  the  cube  root 
first. 


EXAMPLES  FOR  PRACTICE. 
116.      Extract  the 

(a)  Fourth  root  of  100.  f  (a)  3. 16227 

(b)  Fourth  root  of  3,049,800,625.       Ans.  j  (b)  235. 
(f)  Sixth  root  of  9,474,296,896.  (V)  46. 


42  ARITHMETIC.  §  2 


BATIO. 

117.  Suppose  that  it  is  desired  to  compare  two  num- 
bers, say  20  and  4.     If  we  wish  to  know  how  many  times 
larger  20  is  than  4,  we  divide  20  by  4  and  obtain  5  for  the 
quotient;  thus,  20 -f- 4  =  5.     Hence,   we  say  that   20   is  5 
times  as  large  as  4,  i.  e. ,  20  contains  5  times  as  many  units 
as  4.     Again,  suppose  we  desire  to  know  what  part  of  20  is 
4.     We  then  divide  4  by  20  and  obtain  J;  thus,  4 -=-20  —  |, 
or  .2.     Hence,  4  is  ^  or  .2  of  20.     This  operation  of  com- 
paring two  numbers  is  termed  finding  the  ratio  of  the  two 
numbers.     Ratio,  then,  is  a  comparison.     It  is  evident  that 
the  two  numbers  to  be  compared  must  be  expressed  in  the 
same  unit;  in  other  words,  the  two  numbers  must  both  be 
abstract  numbers  or  concrete  numbers  of  the  same  kind. 
For  example,  it  would  be  absurd  to  compare  20  horses  with 
4  birds,  or  20  horses  with  4.     Hence,  ratio  may  be  denned 
as  a  comparison  between  two  numbers  of  the  same  kind. 

118.  A  ratio  may  be  expressed  in  three  ways;  thus,  if 
it  is  desired  to  compare  20  and  4,  and  express  this  compari- 

20 
son  as  a  ratio,  it  may  be  done  as  follows :  20  -f-  4,  20  : 4,  or  — . 

All  three  are  read  the  ratio  of  20  to  4.     The  ratio  of  4  to  20 

4 
would  be  expressed  thus:    4^20,  4:20,  or  — .     The  first 

yv\/ 

method  of  expressing  a  ratio,  although  correct,  is  seldom  or 
never  used ;  the  second  form  is  the  one  of tenest  met  with, 
while  the  third  is  rapidly  growing  in  favor,  and  is  likely  to 
supersede  the  second.  The  third  form,  called  the  fractional 
form,  is  preferred  by  modern  mathematicians,  and  possesses 
great  advantages  to  students  of  algebra  and  of  higher  mathe- 
matical subjects.  The  second  form  seems  to  be  better 
adapted  to  arithmetical  subjects,  and  is  the  one  we  shall 
ordinarily  adopt.  There  is  still  another  way  of  expressing 
a  ratio,  though  seldom  or  never  used  in  the  case  of  a  simple 
ratio  like  that  given  above.  Instead  of  the  colon,  a  straight 
vertical  line  is  used ;  thus,  20  I  4. 


§  2  ARITHMETIC.  43 

119,  The  terms  of  a  ratio  are  the  two  numbers  to  be 
compared  ;  thus,  in  the  above  ratio,  20  and  4  are  the  terms. 
When  both  terms  are  considered  together  they  are  called  a 
couplet  ;    when    considered    separately,    the    first    term   is 
called  the  antecedent,  and  the  second  term,   the  conse- 
quent.    Thus,  in  the  ratio  20  :  4,  20  and  4  form  a  couplet, 
and  20  is  the  antecedent,  and  4,  the  consequent. 

120.  A  ratio  may  be  direct  or  inverse.     The  direct 
ratio  of  20  to  4  is  20  :  4,  while  the  inverse  ratio  of  20  to  4  is 
4  :  20.     The  direct  ratio  of  4  to  20  is  4  r-20,  and  the  inverse 
ratio   is   20  :  4.     An    inverse    ratio    is    sometimes   called  a 
reciprocal  ratio.     The  reciprocal  of  a  number  is  1  divided 

by  the  number.     Thus,  the  reciprocal  of  17  is  —  ;  of  f  is 

1  -f-  1  =  |  ;   i.e.,  the  reciprocal  of  a  fraction  is  the  frac- 
tion inverted.     Hence,  the  inverse  ratio  of  20  to  4  may  be 

expressed  as  4  :  20  or  as  —  -  :  —  .     Both  have  equal  values  ;  for, 
/cO    4 


121.  The  term  vary  implies  a  ratio.     When  we  say  that 
two  numbers  vary  as  some  other  two  numbers,  we  mean 
that  the  ratio  between  the  first  two  numbers  is  the  same  as 
the  ratio  between  the  other  two  numbers. 

122.  The  value  of  a  ratio  is  the  result  obtained  by  per- 
forming the  division   indicated.     Thus,   the  value  of   the 
ratio  20  :  4  is  5  ;  it  is  the  quotient  obtained  by  dividing  the 
antecedent  by  the  consequent. 

123.  By  expressing  the  ratio  in  the  fractional  form,  for 

20 
example,  the  ratio  of  20  to  4  as  —  ,  it  is  easy  to  see,  from 

the  laws  of  fractions,  that  if  both  terms  be  multiplied  or 
both  divided  by  the  same  number  it  will  not  alter  the  value 
of  the  ratio.  Thus, 

20        20X5        100          ,20        20  -=-4        5 


.  _     _  _  •-  -  — 

4X5      '  "20  4    "  '    4-J-4      "  1' 


ARITHMETIC. 


It  is  also  evident,  from  the  laws  of  fractions,  that 
multiplying1  the  antecedent  or  dividing  the  consequent  mul- 
tiplies the  ratio,  and  dividing  the  antecedent  or  multiplying 
the  consequent  divides  the  ratio. 

125.  When  a  ratio  is  expressed  in  words,  as  the  ratio  of 
20  to  4,  the  first  number  named  is  always  regarded  as  the 
antecedent  and  the  second  as  the  consequent,  without  regard 
to  whether  the  ratio  itself  is  direct  or  inverse.  When  not 
otherwise  specified,  all  ratios  are  understood  to  be  direct. 
To  express  an  inverse  ratio  the  simplest  way  of  doing  it  is 
to  express  it  as  if  it  were  a  direct  ratio,  with  the  first  num- 
ber named  as  the  antecedent,  and  then  transpose  the  ante- 
cedent to  the  place  occupied  by  the  consequent  and  the 
consequent  to  the  place  occupied  by  the  antecedent  ;  or  if 
expressed  in  the  fractional  form,  invert  the  fraction.  Thus, 
to  express  the  inverse  ratio  of  20  to  4,  first  write  it  20  :  4, 

and  then,  transposing  the  terms,  as  4  :  20  ;  or  as  —  ,   and 

4 
then  inverting,  as  —  .     Or,  the  reciprocals  of  the  numbers 


may  be  taken,  as  explained  above. 
transpose  its  terms. 


To  invert  a  ratio  is  to 


EXAMPLES  FOR  PRACTICE. 

126.     What  is  the  value  of  the  ratio  of: 


(a) 

w 
w 

(<0 

w 


(h) 

(0 

01 

(*) 


98  :  49  ? 

$45  :  $9  ? 
6J:f? 
3.5:4.5? 
The  inverse 
The  inverse 
The  inverse 
The  inverse 
The  ratio  of 
The  ratio  of 
The  ratio  of 
The  ratio  of 


ratio  of  76  to  19  ? 

ratio  of  49  to  98  ? 

ratio  of  18  to  24  ? 

ratio  of  9  to  15  ? 

10  to  3,  multiplied  by  3  ? 

35  to  49,  multiplied  by  7  ? 

18  to  64,  divided  by  9  ? 

14  to  28,  divided  by  5  ? 


Ans. 


(«)      2. 


(c) 
(</) 
(') 

(/) 


(0 

(/) 
(*) 

(0 


.7ft. 

*. 
2. 

H. 
If. 

10. 

5. 
A- 

TV 


127.     Instead  of  expressing  the  value  of  a  ratio  by  a 
single  number  as  above,  it  is  customary  to  express  it  by 


§  2  ARITHMETIC.  45 

means  of  another  ratio  in  which  the  consequent  is  1.  Thus, 
suppose  that  it  is  desired  to  find  the  ratio  of  the  weights  of  two 
pieces  of  iron,  one  weighing  45  pounds  and  the  other  weigh- 
ing 30  pounds.  The  ratio  of  the  heavier  to  the  lighter  is 
then  45  :  30,  an  inconvenient  expression.  Using  the  frac- 

45 

tional  form,  we  have  — .     Dividing  both  terms  by  30,  the 
ou 

consequent,  we  obtain  -^  or  1^ :  1.  This  is  the  same  result 
as  obtained  above,  for  1^  -~  1  =  1^,  and  45  -r-  30  =  1^. 


A  ratio  may  be  squared,  cubed,  or  raised  to  any 
power,  or  any  root  of  it  may  be  taken.  Thus,  if  the  ratio 
of  two  numbers  is  105  : 63,  and  it  is  desired  to  cube  this 
ratio,  the  cube  may  be  expressed  as  1053 :  633.  That  this  is 
correct  is  readily  seen ;  for,  expressing  the  ratio  in  the  frac- 

..  'f  105  ..    /105\3       1053 

tional  form,  it  becomes  — —,  and  the  cube  is  I  — — -  I  —  -T^- 

bo  \  00  /  00 

=  1053 : 633.  Also,  if  it  is  desired  to  extract  the  cube  root 
of  the  ratio  1053 : 633,  it  may  be  done  by  simply  dividing 
the  exponents  by  3,  obtaining  105  :  63.  „  This  may  be  proved 
in  the  same  way  as  in  the  case  of  cubing  the  ratio.  Thus, 

105* :  63'  =(m\  and  fW=™        105: 63. 
\  bo  /  \  bo  /  bo 


129.     Since    l=jj-\    =  (^  \ ,    it    follows    that    1053:633 

=  53 : 33  (this  expression  is  read,  the  ratio  of  105  cubed 
to  63  cubed  equals  the  ratio  of  5  cubed  to  3  cubed),  and, 
hence,  that  the  antecedent  and  consequent  may  both  be 
multiplied  or  both  divided  by  the  same  number,  irrespec- 
tive of  any  indicated  powers  or  roots,  without  altering  the 
value  of  the  ratio.  Thus,  242:182  =  42:32.  For,  perform- 
ing the  operations  indicated  by  the  exponents,  242  ==  576 
and  182  =  324.  Hence,  576:324  =  If  or  If  :1.  Also,  42 
=  16  and  32  =  9;  hence,  16 : 9  =  If  or  If :  1,  the  same 

94.2          /94.\2          /4A3 

result     as     before.       Also,    24* :  18'  =  j|i  =  (jf j    =  (|) 
=4!  =  4°  =  3'. 


46  ARITHMETIC.  §  2 

The  statement  may  be  proved  for  roots  in  the  same 
manner.  Thus  VW  :  &W  =  &4?  :  ^31  For,  the  V 243 
=  24  and  VW  =  18;  and,  24:18  =  1£  or  1J :  1.  Also, 
^?  =  4  and  VW  =  3 ;  4  :  3  =  1 J  or  1 J :  1. 

If  the  numbers  composing  the  antecedent  and  consequent 
have  different  exponents,  or  if  different  roots  of  those  num- 
bers are  indicated,  the  operations  above  described  cannot  be 
performed.  This  is  evident;  for,  consider  the  ratio  of 
42 : 83.  When  expressed  in  the  fractional  form  it  becomes 

42  /4\2  /4\3 

•TJ,  which  cannot  be  expressed  either  as  I  -J   or  as  1-1  ,  and, 

hence,  cannot  be  reduced  as  described  above. 


PROPORTION. 

130.  Proportion  is  an  equality  of  ratios,  the  equality 
being-  indicated  by  the  double  colon  ( : : )  or  by  the  sign  of 
equality  (=).    Thus,  to  write  in  the  form  of  a  proportion  the 
two  equal  ratios,  8 : 4  and  6 : 3,  which  both  have  the  same 
value,  2,  we  may  employ  one  of  the  three  following  forms : 

8:4  ::  6:3         (1) 
8:4  =  6:3         (2) 

8  -  6  (3} 

4  -  3 

131.  The  first  form  is  the  one  most  extensively  used,  by 
reason  of  its  having  been  exclusively  employed  in  all  the 
older  works  on  mathematics.     The  second  and  third  forms 
are  being  adopted  by  all  modern  writers  on  mathematical 
subjects,  and,  in  time,  will  probably  entirely  supersede  the 
first  form.     In  this  subject  we  shall  adopt  the  second  form, 
unless  some  statement  can  be  made  clearer  by  using  the 
third  form. 

132.  A  proportion  may  be  read  in  two  ways.     The  old 
way  to  read  the  above  proportion  was — 8  is  to  4  as  6  is  to  3; 
the  new  way  is — the  ratio  of  8  to  4  equals  the  ratio  of  6  to  3. 
The  student  may  read  it  either  way,  but  we  recommend  the 
latter. 


§  2  ARITHMETIC.  47 

133.  Each  ratio  of  a  proportion  is  termed  a  couplet. 
In  the  above  proportion,  8  : 4  is  a  couplet,  and  so  is  0  :  3. 

134.  The  numbers  forming  the  proportion  are  called 
terms ;  and  they  are  numbered  consecutively  from  left  to 

right,   thus :  first  second     third  fourth 

8:4  =  6:3 

Hence,  in  any  proportion,  the  ratio  of  the  first  term  to 
the  second  term  equals  the  ratio  of  the  third  term  to  the 
fourth  term. 

135.  The  first  and  fourth  terms  of  a  proportion  are 
called  the  extremes,  and  the  second  and  third  terms,  the 
means.     Thus,  in  the  foregoing  proportion,  8  and  3  are  the 
extremes  and  4  and  6  are  the  means. 

136.  A  direct  proportion  is  one  in  which  both  coup- 
lets are  direct  ratios. 

137.  An  inverse  proportion  is  one  which  requires  one 
of  the  couplets  to  be  expressed  as  an  inverse  ratio.     Thus, 
8  is  to  4  inversely  as  3  is  to  6  must  be  written  8:4  =  6:3; 
i.  e.,  the  second  ratio  (couplet)  must  be  inverted. 

138.  Proportion  forms  one  of  the  most  useful  sections 
of  arithmetic.    In  our  grandfathers'  arithmetics,  it  was  called 
"The  rule  of  three." 

139.  Rule  I. — In   any  proportion,    tJie  product  of  the 
extremes  equals  the  product  of  tJie  means. 

Thus,  in  the  proportion, 

17:51  =  14:42. 
17X42  =  51  X  14,  since  both  products  equal  714. 

140.  Rule  II. —  The  product  of  the  extremes  divided  by 
either  mean  gives  the  other  mean. 

EXAMPLE. — What  is  the  third  term  of  the  proportion  17  :  51  —     :  42  ? 
SOLUTION.— Applying  rule  II,  17X42  =  714,  and  714  -H  51  =  14.  Ans. 

141.  Rule  III. —  The  product  of  the  means  divided  by 
either  extreme  gives  the  other  extreme. 

EXAMPLE. — What  is  the  first  term  of  the  proportion     :  51  —  14  :  42  ? 
SOLUTION.— Applying  rule   III,  51  X  14  =  714,    and   714  -r-  42  =  17. 

Ans. 


48  ARITHMETIC  §  2 

142.  When  stating  a  proportion  in  which  one  of  the 
terms  is  unknown,  represent  the  missing  term  by  a  letter, 
as  x.     Thus,  the  last  example  would  be  written, 

x:5l  =  14:42 

and  for  the  value  of  x  we  have  x  —  —  —  —  =  17. 

42 

143.  If  the  same  operations  (addition  and  subtraction 
excepted)  be  performed  upon  all  the  terms  of  a  proportion, 
the  proportion  is  not  thereby  destroyed.     In  other  words,  if 
all  the  terms  of  a  proportion  be  (1)  multiplied  or  (2)  divided 
by  the  same  number;  (3)  if  all  the  terms  be  raised  to  the 
same  power;  (4)  if  the  same  root  of  all  the  terms  be  taken, 
or  (5)  if  both  couplets  be  inverted,  the  proportion  still  holds. 
We  will  prove  these  statements  by  a  numerical  example,  and 
the  student  can  satisfy  himself  by  other  similar  ones.     The 
fractional  form  will  be  used,  as  it  is  better  suited  to  the  pur- 
pose.   Consider  the  proportion  8  :  4  =  6  :  3.     Expressing  it  in 

8         fi 
the  third  form,  it  becomes  -  —  -.     What  we  are  to  prove  is 

that  if  any  of  the  five  operations  enumerated  above  be 
performed  upon  all  the  terms  of  the  proportion,  the  first 
fraction  will  still  equal  the  second  fraction. 

8  y  7 
1.     Multiplying  all  the  terms  by  any  number,  say  7,  -r  —  = 

6X7          56        42      AT       56  1   42 

°r  28  =  21"  W  28  evldently  ecluals  gj»  smce  the 


value  of  either  ratio  is  2,  and  the  same  is  true  of  the  original 
proportion. 

o    .    iy 

2.  Dividing  all  the  terms  by  any  number,   say  7,       ' 

-  §±£  or  I  =  |.     But  ^  =  2,  and  ^  =  2also,  the 

same  as  in  the  original  proportion. 

3.  Raising  all  the  terms  to  the  same  power,  say  the  cube, 

83         H3  83         /8\3 

—  —  _      This  is  evidently  true,  since  -p  =  f-j  =  23  =  8, 

and  J  =  (|)  =  8-  =  8  also,         . 


§  2  ARITHMETIC.  49 

4.  Extracting  the  same  root  of  all  the  terms,  say  the  cube 
root,  -p=  =  -—==.     It  is  evident   that  this   is  likewise  true, 

-T=  =  y  o  =    &%  also. 

5.  Inverting  both  couplets,  —  =  -,  which  is  true,  since 

o        o 

both  equal  £. 

144.  If  both  terms  of  either  couplet  be  multiplied  or 
both  divided  by  the  same  number,   the  proportion  is  not 
destroyed.     This   should   be   evident   from   the   preceding 
article,  and  also  from  Art.  123.     Hence,  in  any  proportion, 
equal  factors  may  be  canceled  from  the  terms  of  a  couplet, 
before    applying   rule    II   or   III.      Thus,    the   proportion 
45  : 9  =  x\  7.1,  we  may  divide  both  terms  of  the  first  coup- 
let by  9  (that  is,  cancel  9  from  both  terms),  obtaining  5 : 1 
=  x\  7.1,  whence  x  =  7.1  X  5-^1  =  35.5.     (See  Art.  129.) 

145.  The  principle  of  all  calculations  in  proportion  is 
this :   TJiree  of  the  terms  are  always  given,  and  the  remain- 
ing one  is  to  be  found. 

146.  EXAMPLE. — If  4  men  can  earn  $25 -in  one  week,  how  much 
can  12  men  earn  in  the  same  time  ? 

SOLUTION. — The  required  term  must  bear  the  same  relation  to  the 
given  term  of  the  same  kind,  as  one  of  the  remaining  terms  bears  to 
the  other  remaining  term.  We  can  then  form  a  proportion  by  which 
the  required  term  may  be  found. 

The  first  question  the  student  must  ask  himself  in  every  calculation 
by  proportion  is: 

44  What  is  it  I  want  to  find  ?-" 

In  this  case  it  is  dollars.  We  have  two  sets  of  men,  one  set  earning 
$25,  and  we  want  to  know  how  many  dollars  the  other  set  earns.  It  is 
evident  that  the  amount  12  men  earn  bears  the  same  relation  to  the 
amount  4  men  earn  as  12  men  bear  to  4  men.  Hence,  we  have  the 
proportion,  the  amount  12  men  earn  is  to  $25  as  12  men  are  to  4  men, 
or,  since  either  extreme  equals  the  product  of  the  means  divided  by 
the  other  extreme,  we  have 

The  amount  12  men  earn  :  $25  ::  12  men  :  4  men, 

<jj>O  PJ  vx  1  2 

or  the  amount  12  men  earn  =  —  -j =  $75.     Ans. 


50  ARITHMETIC.  §  2 

Since  it  matters  not  which  place  .r,  or  the  required  term,  occupies, 
the  problem  could  be  stated  in  any  of  the  following  forms,  the  value  of 
x  being  the  same  in  each : 

(a)  $25  :  the  amount  12  men  earn  =  4  men  :  12  men ;  or  the  amount 

cj9K  V  1 2 

12  men  earn  = ^ ,  or  $75,  since  either  mean  equals  the  product 

of  the  extremes  divided  by  the  other  mean. 

(b)  4  men  :  12  men  =  $25  :  the  amount  that  12  men  earn ;  or  the 

(COR  *y  1  O 

amount  that  12  men  earn  =  —— ^ ,  or  $75,  since  either  extreme 

equals  the  product  of  the  means  divided  by  the  other  extreme. 

(c)  12  men  :  4  men  =  the  amount  12  men  earn  :  $25 ;  or  the  amount 

that  12  men  earn  =  — — -? ,  or  $75,   since  either  mean  equals  the 

product  of  the  extremes  divided  by  the  other  mean. 

147.  If  the  proportion  is  an  inverse  one,  first  form  it  as 
though  it  were  a  direct  proportion,  and  then  invert  one  of 
the  couplets. 

EXAMPLES  FOR  PRACTICE. 

148.  Find  the  value  of  x  in  each  of  the  following: 


(a)  $16  :  $64  ::  x :  $4. 

(£)  x:  85::  10:  17. 

(c)  24 :  x ::  15  :  40. 

(d)  18 :  94 ::  2  :  x.  Ans. 
(<?)  $75  :  $100  =  * :  100. 

(/)  15  pwt. :  x  =  21 : 10. 

(g)  x  :  75  yd.  =  $15  :  $5. 


(a)  x  =±  $1. 

(b)  x  —  50. 

(c)  x  =  64. 

(d)  *  =  10J. 

(e)  x  -  75. 

(f)  x  =  1\  pwt. 

(g)  x  =  225  yd. 


1.  If  75  pounds  of  lead  cost  $2.10,  what  would  125  pounds  cost  at 
the  same  rate?  Ans.  $3.50. 

2.  If  A  does  a  piece  of  work  in  4  days  and  B  does  it  in  7  days,  how 
long  will  it  take  A  to  do  what  B  does  in  63  days  1  Ans.  36  days. 

3.  The  circumferences  of  any  two  circles  are  to  each  other  as  their 
diameters.     If  the  circumference  of  a  circle  7  inches  in  diameter  is  22 
inches,  what  will  be  the  circumference  of  a  circle  31  inches  in  diameter  ? 

Ans.  97|  inches. 

INVERSE    PROPORTION. 

149.  In  Art.  137,  an  inverse  proportion  was  defined  as 
one  which  required  one  of  the  couplets  to  be  expressed  as  an 
inverse  ratio.  Sometimes  the  word  inverse  occurs  in  the 
statement  of  the  example ;  in  such  cases,  the  proportion  can 


§  2  ARITHMETIC.  51 

be  written  directly,  merely  inverting  one  of  the  couplets. 
But  it  frequently  happens  that  only  by  carefully  studying 
the  conditions  of  the  example,  can  it  be  ascertained  whether 
the  proportion  is  direct  or  inverse.  When  in  doubt,  the 
student  can  always  satisfy  himself  as  to  whether  the  propor- 
tion is  direct  or  inverse  by  first  ascertaining  what  is  required, 
and  stating  the  proportion  as  a  direct  proportion.  Then,  in 
order  that  the  proportion  may  be  true,  if  the  first  term  is 
smaller  than  the  second  term,  the  third  term  must  be  smaller 
than  the  fourth ;  or  if  the  first  term  is  larger  than  the  second 
term,  the  third  term  must  be  larger  than  the  fourth  term. 
Keeping  this  in  mind,  the  student  can  always  tell  whether 
the  required  term  will  be  larger  or  smaller  than  the  other 
term  of  the  couplet  to  which  the  required  term  belongs. 
Having  determined  this,  the  student  then  refers  to  the 
example,  and  ascertains  from  its  conditions  whether  the 
required  term  is  to  be  larger  or  smaller  than  the  other  term 
of  the  same  kind.  If  the  two  determinations  agree,  the  pro- 
portion is  direct,  otherwise  it  is  inverse,  and  one  of  the 
couplets  must  be  inverted. 

1 5O.  EXAMPLE. — If  A's  rate  of  doing  work  is  to  B's  as  5  :  7,  and  A 
does  a  piece  of  work  in  42  days,  in  what  time  will  B  do  it  ? 

SOLUTION. — The  required  term  is  the  number  of  days  it  will  take  B 
to  do  the  work.  Hence,  stating  as  a  direct  proportion, 

5  :  7  =  42  :  x. 

Now,  since  7  is  greater  than  5,  x  will  be  greater  than  42.  But,  refer- 
ring to  the  statement  of  the  example,  it  is  easy  to  see  that  B  works 
faster  than  A ;  hence  it  will  take  B  a  less  number  of  days  to  do  the 
work  than  A.  Therefore,  the  proportion  is  an  inverse  one,  and  should 
be  stated 

5  :  7  =  x  :  42 

from  which  x  =  -~ —  =  30  days.     Ans. 

Had  the  example  been  stated  thus :  The  time  that  A  requires  to  do 
a  piece  of  work  is  to  the  time  that  B  requires,  as  5  :  7 ;  A  can  do  it  in  42 
days,  in  what  time  can  B  do  it?  it  is  evident  that  it  would  take  B  a 
longer  time  to  do  the  work  than  it  would  A ;  hence,  x  would  be  greater 

7  X  42 
than  42,  and  the  proportion  would  be  direct,  the  value  of  x  being  — - — 

=  58.8  davs. 


52  ARITHMETIC.  §  2 

EXAMPLES  FOR  PRACTICE. 

151.  Solve  the  following : 

1.  If  a  pump  which  discharges  4  gal.  of.water  per  min.  can  fill  a  tank 
in  20  hr. ,  how  long  will  it  take  a  pump  discharging  12  gal.  per  min.  to 
fill  it?  Ans.  6f  hr. 

2.  If  a  pump  discharges  90  gal.  of  water  in  20  hr. ,  in  what  time  will 
it  discharge  144  gal.  ?  Ans.  32  hr. 

3.  The  weight  of  any  gas  (the  volume  and  pressure  remaining  the 
same)  varies  inversely  as  the   absolute   temperature.      If  a  certain 
quantity  of  some  gas  weighs  2.927  Ib.  when  the  absolute  temperature 
is  525°,  what  will  the  same  volume  of  gas  weigh  when  the  absolute 
temperature  is  600°,  the  pressure  remaining  the  same  ?    Ans.  2.561+  Ib. 

4.  If  50  cu.  ft.  of  air  weigh  4.2  pounds  when  the  absolute  tempera- 
ture is  562°,  what  will  be  the  absolute  temperature  when  the  same 
volume  weighs  5.8  pounds,  the  pressure  being  the  same  in  both  cases  ? 

Ans.  407°,  very  nearly. 

POWERS    AND    ROOTS    IN    PROPORTION. 

152.  It  was  stated  in  Art.   128  that  a  ratio  could  be 
raised  to  any  power  or  any  root  of  it  might  be  taken.     A 
proportion  is  frequently  stated  in  such  a  manner  that  one  of 
the  couplets  must  be  raised  to  some  power  or  some  root  of 
it  must  be  taken.     In   all   such  cases,   both  terms  of  the 
couplet  so  affected  must  be  raised  to  the  same  power  or  the 
same  root  of  both  terms  must  be  taken. 

153.  EXAMPLE. — Knowing  that  the  weight  of  a  sphere  varies  as 
the  cube  of  its  diameter,  what  is  the  weight  of  a  sphere  6  inches  in 
diameter  if  a  sphere  8  inches  in  diameter  of  the  same  material  weighs 
180  pounds  ? 

SOLUTION. — This  is  evidently  a  direct  proportion.     Hence,  we  write 

63 :  83  =  x  :  180. 

Dividing  both  terms  of  the  first  couplet  by  23  (see  Art.  129) 
33 :  43  =  x  :  180,  or  27  :  64  =  x  :  180; 

27  X  180 
whence,  x  =  — -^- =  75{|  pounds.     Ans. 

UT: 

EXAMPLE. — A  sphere  8  inches  in  diameter  weighs  180  pounds;  what 
is  the  diameter  of  another  sphere  of  the  same  material  which  weighs 
75}|  pounds  ? 

SOLUTION. — Since  the  weights  of  any  two  spheres  are  to  each  other 
as  the  cubes  of  their  diameters,  we  have  the  proportion 
180  :  75||  =  8s :  .*». 


§  2  ARITHMETIC.  53 

The  required  term,  x,  must  be  cubed,  because  the  other  term  of  the 
couplet  is  cubed  (see  Art.  152).     But,  83  =  512;  hence, 


180  :  75*£  =  512  :  .*»,  or  jr8  =        *  =  216; 

J.OU 

whence,  x  =    ^216  =  6  inches.     Ans. 

154.  Since  taking  the  same  root  of  all  the  terms  of  a 
proportion  does  not  change  its  value  (Art.  143),  the  above 
example  might  have  been  solved  by  extracting  the  cube  root 
of  all  the  numbers,  thus  obtaining  ^180  :  ^75|f  =  8:*; 


:      ~  K  64 


=  8  X  f  =  6  inches.     The  process,  however,  is  longer  and  is 
not  so  direct,  and  the  first  method  is  to  be  preferred. 

155.  If  two  cylinders  have  equal  volumes,  but  different 
diameters,  the  diameters  are  to  each  other  inversely  as  the 
square  roots  of  their  lengths.  Hence,  if  it  is  desired  to  find 
the  diameter  of  a  cylinder  that  is  to  be  15  inches  long,  and 
which  shall  have  the  same  volume  as  one  that  is  9  inches 
in  diameter  and  12  inches  long,  we  write  the  proportion 
9  :  x 


Since  neither  12  nor  15  are  perfect  squares,  we  square  all 
the  terms  (Arts.  154  and  143)  and  obtain 

81  :  a*  =  15  :  12;  whence,  x2  =  81.X,12  =  64.8, 

lo 

and     x  =    V64.8  =  8.05     inches  =  diameter     of     15-inch 
cylinder. 

EXAMPLES  FOR  PRACTICE. 

156.      Solve  the  following  examples: 

1.  The  intensity  of  light  varies  inversely  as  the  square  of  the  dis- 
tance from  the  source  of  light.     If  a  gas  jet  illuminates  an  object  30 
feet  away  with   a  certain  distinctness,  how  much  brighter  will  the 
object  be  at  a  distance  of  20  feet  ?  Ans.  2J  times  as  bright. 

2.  In  the  last  example,  suppose  that  the  object  had  been  40  feet 
from  the  gas  jet;  how  bright  would  it  have  been,  compared  with  its 
brightness  at  30  feet  from  the  gas  jet  ?  Ans.  T9¥  as  bright. 

3.  When  comparing  one  light  with  another,  the  intensities  of  their 
illuminating  powers  vary  as  the  squares  of  their  distances  from  the 

1-8 


54  ARITHMETIC.  §  2 

source.  If  a  man  can  just  distinguish  the  time  indicated  by  his  watch, 
50  feet  from  a  certain  light,  at  what  distance  could  he  distinguish  the 
time  from  a  light  3  times  as  powerful  ?  Ans.  86.6+  feet. 

4.  The  quantity  of  air  flowing  through  a  mine  varies  directly  as  the 
square  root  of  the  pressure.     If  60,000  cubic  feet  of  air  flow  per  min- 
ute when  the  pressure  is  2.8  pounds  per  square  foot,  how  much  will 
flow  when  the  pressure  is  3.6  pounds  per  square  foot  ? 

Ans.  68,034  cu.  ft.  per  min.,  nearly. 

5.  In  the  last  example,  suppose  that  70,000  cubic  feet  per  minute 
had  been  required;    what  would  be  the  pressure  necessary  for  this 
quantity?  Ans.  3.81+  Ib.  per  sq.  ft. 


CAUSES  AKD  EFFECTS. 

157.  Many  examples  in  proportion  may  be  more  easily 
solved  by  using  the  principle  of  cause  and  effect.     That 
which  may  be  regarded  as  producing  a  change  or  alteration 
in  something,  or  as  accomplishing  something,  may  be  called 
a  cause,  and  the  change  or  alteration,  or  thing  accomplished, 
as  the  effect. 

158.  Like  causes  produce  like  effects.     Hence,  when  two 
causes  of  the  same  kind  produce  two  effects  of  the  same 
kind,  the  ratio  of  the  causes  equals  the  ratio  of  the  effects ; 
in  other  words  the  first  cause  is  to  the  second  cause  as  the 
first  effect  is  to  the  second  effect.     Thus,  in  the  question — 
if  3  men  can  lift  1,400  pounds,  how  many  pounds  can  7 
men  lift? — we  call  3  men  and  7  men  the  causes  (since  they 
accomplish  something,  viz.,  the  lifting  of  the  weight),  the 
number  of  pounds  lifted,  viz.,  1,400  pounds  and  x  pounds, 
are  the  effects.    If  we  call  3  men  the  first  cause,  1,400  pounds 
is  the  first  effect ;  7  men  is  the  second  cause,  and  x  pounds 
is  the  second  effect.      Hence,  we  may  write 

1st  cause      2d  cause  1st  effect      2d  effect 

3        :'        7        =       1,400       :       x 

7  v  1  4-00 

whence  x  =  i^     -  =  3,266|  pounds. 
o 

159.  The  principle  of  cause  and   effect  is   extremely 
useful  in  the  solution  of  examples  in  compound  proportion, 
as  we  shall  now  show. 


§  2  ARITHMETIC.  55 

COMPOUND  PROPORTION 

16O.  All  the  cases  of  proportion  so  far  considered  have 
been  cases  of  simple  proportion";  i.  e.  ,  each  term  has  been 
composed  of  but  one  number.  There  are  many  cases,  how- 
ever, in  which  two  or  all  the  terms  have  more  than  one 
number  in  them  ;  all  such  cases  belong  to  compound  pro- 
portion. In  all  examples  in  compound  proportion,  both 
causes  or  both  effects  or  all  four  consist  of  more  than  two 
numbers.  We  will  illustrate  this  by  an 

EXAMPLE.  —  If  40  men  earn  $1,280  in  16  days,  how  much  will  36  men 
earn  in  31  days? 

SOLUTION.  —  Since  40  men  earn  something,  40  men  is  a  cause,  and 
since  they  take  16  days  in  which  to  earn  something,  16  days  is  also  a 
cause.  For  the  same  reason  36  men  and  31  days  are  also  causes.  The 
effects,  that  which  is  earned,  are  1,280  dollars  and  x  dollars.  Then,  40 
men  and  16  days  make  up  the  first  cause,  and  36  men  and  31  days 
make  up  the  second  cause.  $1,280  is  the  first  effect,  and  «jLr  is  the  second 
effect.  Hence,  we  write 

1st  cause  2d  cause          1st  effect  2d  effect 
40  36  1  9Sn 

16      :       31  l'm 

Now,  instead  of  using  the  colon  to  express  the  ratio,  we  shall  use  the 
vertical'line  (see  Art  118),  and  the  above  becomes 
40    I    36  _  1  280    I 
16    I    31  -  I'280    I    ~r" 

In  the  last  expression,  the  product  of  all  the  numbers  included 
between  the  vertical  lines  must  equal  the  product  of  all  the  numbers 
without  them  ;  i.  e.,  36  X  31  X  1,280  =  40  X  16  X  x  . 

2 

=  $2,23,   Ans.          , 


161.  The  above  might  have  been  solved  by  canceling' 
factors  of  the  numbers  in  the  original  proportion.  For,  it 
any  number  within  the  lines  has  a  factor  common  to  any 
number  without  the  lines,  that  factor  may  be  canceled  from 
both  numbers.  Thus, 

2 
36    _        ft) 

si       ;2^0 

16  is  contained  in  1,280,  80  times.     Cancel  16  and  1,280,  and 
write  80  above  1,280.     40  is  contained  in  80,  2  times.     Cancel 


56 


ARITHMETIC. 


40  and  80,  and  write  2  above  80.  Now,  since  there  are  no  more 
numbers  that  can  be  canceled,  x  —  36  X  31  X  2  =  $2,232,  the 
same  result  as  was  obtained  in  the  preceding  article. 

162.  Rule. —  Write  all  the  numbers  forming  the  first 
cause  in  a  vertical  column,  and  draw  a  vertical  line  ;  on  the 
other  side  of  this  line  write  in   a  vertical  column  all  the 
numbers   forming  the  second   cause.       Write   the  sign   af 
equality  to  the  right  of  the  second  column,  and  on  the  right 
of  this  form  a  third  column  of  the  numbers  composing  the 
first  effect,  drawing  a  vertical  line  to  the  right ;  on  the  other 
side  of  this  line,  write  for  a  fourth   column,  the  numbers 
composing  the  second  effect.      There  must  be  as  many  num- 
bers in  the  second  cause  as  in  the  first  cause,  and  in  the 
second  effect  as  in  the  first  effect ;  hence,    if  any  term  is 
wanting,    write  x  in   its  place.      Multiply  together  all  the 
numbers  within  the  vertical  lines,  and  also  all  those  without 
the  lines  (canceling  previously,    if  possible],  and  divide  the 
product  of  those   numbers   which  do  not  contain  x  by  the 
product  of  the  others  in  which  x  occurs,  and  the  result  will 
be  the  value  of  x. 

163.  EXAMPLE.— If  40  men  can  dig  a  ditch  720  feet  long,  5  feet 
wide,  and  4  feet  deep  in  a  certain  time,  how  long  a  ditch  6  feet  deep 
and  3  feet  wide  could  24  men  dig  in  the  same  time  ? 

SOLUTION. — Here  40  men  and  24  men  are  the  causes,  and  the  two 
ditches  are  the  effects.     Hence, 


24  = 


whence,  x  =  24  X  5  X  4  =  480  feet.     Ans. 


164.  EXAMPLE. — The  volume  of  a  cylinder  varies  directly  as  its 
length  and  directly  as  the  square  of  its  diameter.  If  the  volume  of  a  cylin- 
der 10  inches  in  diameter  and  20  inches  long  is  1 , 570. 8  cubic  inches,  what  is 
the  volume  of  another  cylinder  16  inches  in  diameter  and  24  inches  long  ? 

SOLUTION. — In  this  example,  either  the  dimensions  or  the  volumes 
may  be  considered  the  causes;  say  we  take  the  dimensions  for  the 
causes.  Then,  squaring  the  diameters, 


102 
20 


whence,  x  = 


=  1,570.8 


256X^X1,570.8 
5X100 


100 

or    20 

5 


256 
6 


=  1,570.8 


=  4,825.4976  cubic  inches.    Ans. 


ARITHMETIC. 


165.  EXAMPLE. — If  a  block  of  granite  8ft.  long,  5  ft.  wide,  and 
3  ft.  thick  weighs  7,200  lb.,  what  will  be  the  weight  of  a  block  of 
granite  12  ft.  long,  8  ft.  wide,  and  5  ft.  thick  ? 

SOLUTION. — Taking  the  weights  as  the  effects,  we  have 
4 


=  7,200 


x,  or  x  -  4  X  7,200  =  28,800  pounds,     Ans. 


166.  EXAMPLE. — If  12  compositors  in  30  days  of  10  hours  each 
set  up  25  sheets  of  16  pages  each,  32  lines  to  the  page,  in  how  many 
days  8  hours  long  can  18  compositors  set  up,  in  the  same  type,  64 
sheets  of  12  pages  each,  40  lines  to  the  page  ? 

SOLUTION.  —Here  compositors,  days,  and  hours  compose  the  causes, 
and  sheets,  pages,  and  lines  the  effects.     Hence, 
3         2          3         2 


10 


x    = 


Z,  or  x  =  3  X  10  X  2  =  60  days.     Ans. 


167.  In  examples  stated  like  that  in  Art.  164,  should 
an  inverse  proportion  occur,  write  the  various  numbers  as 
in  the  preceding  examples,  and  then  transpose  from  one 
side  of  the  vertical  line  to  the  other  side  those  numbers 
which  are  said  to  vary  inversely. 

EXAMPLE.  —  The  centrifugal  force  of  a  revolving  body  varies  directly 
as  its  weight,  as  the  square  of  its  velocity,  and  inversely  as  the  radius 
of  the  circle  described  by  the  center  of  the  body.  If  the  centrifugal 
force  of  a  body  weighing  15  pounds  is  187  pounds  when  the  body 
revolves  in  a  circle  having  a  radius  of  12  inches,  with  a  velocity  of  20 
feet  per  second,  what  will  be  the  centrifugal  force  of  the  same  body 
when  the  radius  is  increased  to  18  inches  and  the  speed  is  increased  to 
24  feet  per  second  ? 

SOLUTION.  —  Calling  the  centrifugal  force  the  effect,  we  have 


15 
202 
12 


15 

242  = 
18 


187 


Transposing  12  and  18  (since  the  radii  are  to  vary  inversely)  and  squar- 
ing 20  and  24, 


25 


=187 


12 


12  X  2  * 


=  179-52 


Ans. 


58  ARITHMETIC.  §  2 

•* 

EXAMPLES  FOR  PRACTICE. 

168,      Solve  the  following  by  compound  proportion: 

1.  If  12  men  dig  a  trench  40  rods  long  in  24  days  of  10  hours  each, 
how  many  rods  can  16  men  dig  in  18  days  of  9  hours  each  ? 

Ans.  36  rods. 

2.  If  a  piece  of  iron  7  feet  long,  4  inches  wide,  and  6  inches  thick 
weighs  600  pounds,  how  much  will  a  piece  of  iron  weigh  that  is  16  feet 
long,  8  inches  wide,  and  4  inches  thick  ?  Ans.  l,828f  Ib. 

3.  If  24  men  can  build  a  wall  72  rods  long,  6  feet  wide,  and  5  feet 
high  in  60  days  of  10  hours  each,  how  many  days  will  it  take  32  men 
to  build  a  wall  96  rods  long,  4  feet  wide,  and  8  feet  high,  working 
8  hours  a  day  ?  Ans.  80  days. 

4.  The  horsepower  of  an  engine  varies  as  the  mean  effective  pres- 
sure, as  the  piston  speed,  and  as  the  square  of  the  diameter  of  the 
cylinder.     If    an   engine    having   a  cylinder    14    inches  in   diameter 
develops  112  horsepower  when  the  mean  effective  pressure  is  48  pounds 
per  square  inch  and  the  piston  speed  is  500  feet  per  minute,  what  horse- 
power will   another  engine   develop   if  the  cylinder  is   16  inches  in 
diameter,  piston  speed  is  600  feet  per  minute,  and  mean  effective  pres- 
sure is  56  pounds  per  square  inch  ?  Ans.  204.8  horsepower. 

5.  Referring  to  the  example  in  Art.  164,  what  will  be  the  volume 
of  a  cylinder  20  inches  in  diameter  and  24  inches  long  ?   , 

Ans.  7,539.84  cubic  inches. 

6.  Knowing  that  the  product  of  3X5X7X9  is  945,  what  is  the 
product  of  6  X  15  X  14  X  36  ?  Ans.  45,360. 


ELEMENTARY  ALGEBRA 

AND 

TRIGONOMETRIC  FUNCTIONS. 

ELEMENTS    OF  ALGEBRA. 


USE  OF  LETTERS. 

1.  In  arithmetic,  numbers  are  represented  by  the  figures 
1,  2,  3,  4,  etc.  There  is  no  reason,  however,  why  numbers 
may  not  be  represented  by  other  symbols,  such  as  letters,  if 
rules  are  provided  for  their  use. 

3.  In  algebra,  numbers  are  represented  by  both  figures 
and  letters.  It  will  be  seen  later  that  the  use  of  letters 
often  simplifies  the  solution  of  examples  and  shortens  cal- 
culations. 

3.  The  principal  advantage  of  letters  is  that  they  are 
general  in  their  meaning.     Thus,  unlike  figures,  the  letter 
a  does  not  stand  for  the  number  1,  the  letter  b  for  2,  c  for  3, 
etc. ,  but  any  letter  may  be  taken  to  represent  any  number, 
it  being  only  necessary  that  a  letter  shall  always  stand  for 
the  same  number  in  the  same  example. 

4.  To    illustrate    this    difference    between    letters    and 
figures,  we  may  take  an  example,  as  follows :    If  a  farmer 
exchanges  20  bushels  of  oats,  worth  40  cents  per  bushel, 

§  3 


2  ELEMENTARY  ALGEBRA  AND  §  3 

for  8  bushels  of  wheat,  what  is  the  price  of  the  wheat  per 
bushel  ?  A  rule  for  solving  this  problem,  and  others  like  it, 
would  be  as  follows:  Multiply  the  number  of  bus!  ids  of  oats 
by  the  price  per  busliel,  and  divide  the  result  by  the  number 
of  bushels  of  wheat. 

This  rule  is  general,  because  it  tells  us  what  to  do  with 
the  number  of  bushels  and  with  the  prices  of  the  oats  and 
the  wheat,  whatever  they  may  be. 

A  more  concise  way  of  stating  this  rule  is  to  use  letters  in 
the  same  manner  as  in  formulas.  .Thus  : 

Let  a  =  number  of  bushels  of  oats; 

b  =  price  per  bushel  of  oats; 

c  =  number  of  bushels  of  wheat  ; 

d  =  price  per  bushel  of  wheat. 
Then,  according  to  the  rule, 

bushels  of  oats  X  price  per  bushel 
-  =-  —  r—  -,  —  £  —  r~  -  =  price  of  wheat, 
bushels  ot  wheat 


or 


In  the  example  in  question,  a  =  20,  b  =  40,  and  c  =.  8. 
Hence,  writing  for  a,  b,  and  c  their  values,  20,  40,  and  8,  d, 

20  X  40 
the  price  per  bushel  of  wheat  =  —  —   -  —  100.      Here  the 


20X40  . 

expression  •  —       -  corresponds  to  -   —  ,  but  this  difference 

o  C 

is  to  be  noticed:  -          -  applies  only  to  this  example,  and 

o 

by  performing  the  operations  indicated  only  one  answer  can 
be  obtained,  while  -  -  is  general  in  its  application,  in  the 

same  way  that  the  rule  previously  given  is  general.  That 
is,  while  a,  b,  and  c  stand  for  the  numbers  20,  40,  and  8  in 
this  example,  they  may  stand  for  other  numbers  in  another 
example;  hence,  by  writing  their  values  in  place  of  the 
letters,  and  by  performing  the  operations  indicated,  the 
answer  to  any  example  of  the  same  kind  may  be  obtained. 
Consequently,  while  figures  or  combinations  of  figures  always 


§  3  TRIGONOMETRIC  FUNCTIONS.  3 

represent  the  same  numbers,  letters  are  more  general,  and 
may  represent  any  numbers,  according  to  the  conditions  of 
the  example. 

5.  An  equation  is  a  statement  of  equality  between  two 
expressions.      Thus,  x-\-y  —  8  is  an  equation,  and  means 
that  the  sum  of  the  numbers  represented  by  x  and  y  is  equal 
to  8.     Examples  are  solved  in  algebra  by  the  aid  of  equations, 
in  which  numbers  are  represented  both  by  letters  and  by 
figures.     The  following  simple  example  will  give  an  idea  of 
the  method  of  solution: 

EXAMPLE. — If  an  iron  rail  30  feet  long  is  cut  in  two  so  that  one  part 
is  four  times  as  long  as  the  other,  how  long  is  the  shorter  part  ? 
SOLUTION. — Since  any  letter  may  represent  any  number, 
L/et  x  =  the  length  of  the  shorter  part. 
Then,  4  X  -r  (written  4-t)  =  the  length  of  the  longer  part. 
But  the  sum  of  the  two  parts  must  equal  the  total  length,  30  feet. 
Hence,  ,r  +  4.r  =  30. 

Adding  x  and  4^,  5-r  =  30. 

Whence,  dividing  by  5,  x  —  6  feet.     Ans. 

6.  The    student    has    probably   noticed    the    similarity 
between   an   equation    and  a  formula.     All  formulas   are 
equations,  and  the  same  rules  apply  to  both.     An  equation 
is  not  called  a  formula,  however,  unless  it  is  a  statement  of 
a  general  rule. 

7.  Algebra  treats  of  the  equation  and  its  use.      Since 
the  use  of  equations  involves  the  use  of  letters,  it  will  be 
necessary  before  considering  equations  to  take  up  addition, 
subtraction,  multiplication,  etc.  of  expressions  in  which  let- 
ters are  used'. 


KOTATIOX. 

8.  The  term  quantity  is  used  to  designate  any  number 
that  is  to  be  subjected  to  mathematical  processes.  A  quan- 
tity is  strictly  a  concrete  number;  as,  6  books,  5  pounds,  10 
yards.  Symbols  used  to  represent  numbers,  and  expressions 


4  ELEMENTARY  ALGEBRA  AND  §  3 

containing  two  or  more  such  symbols,  as  a,  x,  bd,  10,  (c-\-  12), 
etc.,  are  often  called  quantities,  the  term  being  a  convenient 
one  to  use. 

9.  The  signs  -f  ,  —  ,  X  ,  -r-  are  the  same  in  algebra  as  in 
arithmetic.  The  sign  of  multiplication  x  is  usually  omitted, 
however,  multiplication  being  indicated  by  simply  writing 
the  quantities  together.  Thus,  abc  means  axbxc\  2;rj/  means 
2X-fXj.  Evidently  the  sign  cannot  be  omitted  between 
two  figures,  as  addition  instead  of  multiplication  would  then 
be  indicated.  Thus,  24  means  20  +  4:  instead  of  2  X  4. 


10.  A  coefficient  is   a   figure  or  letter  prefixed   to  a 
quantity  ;  it  shows  how  many  times  the  latter  is  to  be  taken. 
Thus,  in  the  expression  4«,  4  is  the  coefficient  of  a,  and  indi- 
cates that  a  is  to  be  taken  four  times;   that  is,  4#  is  equal  to 
a  -\-a-\-a-\-  a.      When    several    quantities     are     multiplied 
together,  any  of  them  may  be  regarded  as  the  coefficient  of 
the  others.     Thus,  in  6#  ;rj/,  6  is  the  coefficient  of  axy  ;  6#,  of 
xy\  §a.r,  of  y;  etc.     In  general,  however,  when  a  coefficient 
is  spoken  of,  the  numerical  coefficient  only  is  meant,  as  the 
6  above.     When   no   numerical    coefficient   is   written  it  is 
understood  to  be  1.     Thus,  cd  is  the  same  as  \cd. 

11.  The  factors  of  a  quantity  are  the  quantities  which, 
when  multiplied  together,  will  produce  it.     Thus,  2,  3,  and  3 
are  the  factors  of  18,  since  2x3x3  =  18;  2,  a,  and  b  are 
the  factors  of  %ab,  since  2  X  a  X  b  = 


13.  An  exponent  is  a  small  figure  placed  at  the  right 
and  a  little  above  a  quantity  ;  it  shows  how  many  times  the 
latter  is  to  be  taken  as  a  factor.  Thus,  43  =  4x4x4  =  64, 
the  exponent  3  showing  that  the  number  4  is  to  be  used 
three  times  as  a  factor  ;  likewise  a*  =  aaaaa.  Any  quantity 
written  without  an  exponent  is  understood  to  have  the 
exponent  1  ;  thus,  b1  =  b. 

13.  The  difference  between  a  coefficient  and  an  expo- 
nent should  be  clearly  understood.  A  coefficient  multiplies 
the  quantity  which  it  precedes;  it  shows  that  the  quantity  is 


§  3  TRIGONOMETRIC  FUNCTIONS.  5 

to  be  added  to  itself.  Thus,  3a  =  3  X  #  ,  or  a  -f-  a  +  ft.  An 
exponent  indicates  that  a  quantity  is  to  be  multiplied  by 
itself.  Thus,  a9  =  axaXa.  A  more  complete  definition 
of  an  exponent  will  be  given  later. 

14.  A  power  is  the  result  obtained  by  taking  a  quantity 
two  or  more  times  as  a  factor.    For  example,  10  is  the  fourth 
power  of  2,  because  2  multiplied  by  itself  until  it  has  been 
taken  four  times  as  a  factor  produces  16;   a3  is  the  third 
power  of  <z,  because  aXaXa  =  a\ 

15.  A  root  of  a  quantity  is  one  of  its  equal  factors. 
Thus,    2   is   the   root    of   4,    8,    and    1C,    since    2x2  =  4, 
2X2X2  =  8,    and    2x2x2x2  =  16,    2   being  one   of  the 
equal  factors  in  each  case.     In  like  manner,  a  is  a  root  of  #2, 
a3,  a*,  etc.     The  symbol  which  denotes  that  the  second,  or 
square,  root  is  to  be  extracted  is  \/\  it  is  called  the  radical 
sign,  and  the  quantity  under  the  sign  is  called  the  radical. 
For  other  roots  the  same  symbol  is  used,  but  with  a  figure, 
called  the  index  of  the  root,  written  above  it  to  indicate  the 
root.     Thus,  \/a->  Vat  tya,  etc.  signify  the  square  root,  cube 
root,  fourth,  root,  etc.  of  a. 

16.  The  use  of  the  parenthesis,  bracket,  brace,  and  vin- 
culum  is  explained  in  Arithmetic.     These  symbols  are  called 
symbols   of  aggregation,    meaning   that    the    quantities 
enclosed  within  them  are  aggregated,  or  collected,  into  one 
quantity. 

17.  The  terms   of  an  algebraic   expression   are   those 
parts  which  are   connected  by  the  signs  -{-  and  —  .     Thus, 
x~,  —  2,rj/,  and  j/2  are  terms  of  the  expression  x*  —  2xy-\-y2. 
When  a  term  contains  both  figures  and  letters,  the  part  con- 
sisting of  letters  is  called  the  literal  part  of  the  term;  thus, 
xy  is  the  literal  part  of  the  term  ^ 


18.  like  terms  are  those  which  differ  only  in  their 
numerical  coefficients;  all  others  are  unlike  terms.  Thus, 
2##2  and  zab*  are  like  terms;  5ab  and  Sab*  are  unlike  terms, 
because  one  contains  b  and  the  other  P. 


6  ELEMENTARY  ALGEBRA  AND  §  3 

19.     A  monomial  is  an  expression  consisting  of  only  one 
term;  as,  4afc,  3;r2,  2rt.r3,  etc. 


2O.     A   binomial   is   an   expression   consisting   of   two 
terms;  as,  a-\-b^  2#  +  5^,  etc. 


21.  A  trinomial  is  an  expression   consisting  of  three 
terms;    as,   a*  •\-'bab  +  b*,   (a  +  x)*  —*l(a  +  x)y+y>,   etc  ,  the 
expression  (a  +  x)  being  treated  as  one  quantity. 

22.  A  polynomial  is  an  expression  consisting  of  more 
than  one  term.     The  name  is  usually  applied  only  to  an 
expression  consisting  of  four  or  more  terms. 

23.  The  polynomial  a  +  cfb  +  2«8  —  3a*&  —  a5  is  said  to  be 
arranged  according  to  the  increasing  powers  of  a,  because 
the  exponents  of  a  increase  in  each  term  from  left  to  right, 
the  exponent  of  the  first  a  being  1  understood.      (Art.  12.) 
The  polynomial  a*b3  +  <*&*  +  ^a"b  -f-  1  is  arranged  according 
to  the  decreasing  powers  of  b,  the  exponents  of  b  decreasing 
in  order  from  left  to  right. 

24.  The  arrangement  of  the  terms  of  a  polynomial  does 
not  affect  its  value.     Thus,  x*  -\-  ^xy  +/2  has  the  same  value 
as  %xy  +  y*  +  •tf'2,   just   as   2  +  6  +  4  has  the  same  value  as 


READING   ALGEBRAIC    EXPRESSIONS. 

25.  Quantities  like  «,   x,  #2,   etc.  are  read    *'#,"    u.r," 
"  b  square,"  etc.     In  reading  monomials  in  which  multipli- 
cation is  indicated,  the  word  "times "is  not  used.     Thus, 
abc  is  read  " abc  ";  lad*b*  is  read  *'  lad  square  b  cube." 

26.  The    polynomial   a  +  a*b  +  Za3  —  3a*b  —  a*    is    read 
"«,  pltis  a  square  b,  plus  2a  cube,  minus  3a  fourth  ^,  minus 
a  fifth. "    Considerable  care  is  required  when  reading  expres- 
sions containing  polynomials.     Thus,  if  4(#  —  b}  were  read 
' '  4<7  minus  ^, "  the  binomial  ka  —  b  would  be  understood.      It 
should 'be  read  "4  times  a  —  b"  or  "4  times  the  parenthesis 


§  3  TRIGONOMETRIC  FUNCTIONS.  7 

a  minus  b,  "  in  which  case  it  will  be  understood  that  4  multi- 
plies the  whole  quantity  a  —  b,  since  the  word  "times"  is 
not  used  with  monomials.  Again,  m(ii?  -f-  "Zmn  -f-  //a)  and 
;//(///-  -{-  2  mil}  -f-  /f2  should  each  be  so  read  that  there  can  be  no 
doubt  as  to  whether  the  ;/*  is  to  be  multiplied  by  ;//  or  not. 

Let  the  distinction  to  be  made  in  reading  the  following  be 
observed  : 


/ 
A/ 


and  A/  m-\ 
^ 


In  the  first  case,  the  whole  quantity  ;//  +  ;/  *s  divided  by 
x—  y,  and  it  would  be  clear  to  say,  "  the  square  root  of  the 
fraction  m-\-n  over  x—  j/2.  "  In  the  second  case,  where  the 
;/  only  is  divided  by  JIT  —  /*,  it  may  be  read,  "the  square 
root  of  the  quantity,  in  plus  the  fraction  n  over  x—  j'2.  " 
The  word  "quantity"  shows  that  the  square  root  of  the 
whole  expression  is  taken,  and  the  word  "  fraction  "  after 
"  plus  "  shows  that  only  the  n  is  divided  by  x—  /2. 

27.  When  a  polynomial  is  affected  by  an  exponent,  it 
should  be  indicated  clearly.  Thus, 


should  be  read,  "3a  —  d  square,  times  the  square  of  3a  —  d, 
times  the  square  of  Za  —  d  square.  " 

28.  Sometimes  expressions  like  A',  R'\  c',  d"  ',  £T,  <?2, 
etc.  appear  in  formulas  or  elswhere  in  algebraic  problems 
when  it  is  desirable  to  have  the  same  letter  represent  different 
quantities  that  are  similar,  or  correspond  to  one  another. 
The  marks  ',  ",  '",  ,,  2/etc.  serve  to  distinguish  the  letters. 
The  expressions  are  also  used  to  designate  similar  or  corre- 
sponding lines  in  geometrical  figures.  A',  B"  ,  C",  etc. 
are  read  "a  major  prime,  b  major  second,  c  major  tliird," 
etc.;  a',  b"  ,  c'",  etc.  are  read  "a  minor  prime,  b  minor 
second,  c  minor  third,"  etc.;  #,,  /?„,  C^  d^,  etc.  are  read 
"a  minor  sub-one,  b  major  sub-two,  c  major  sub-three, 
d  minor  sub-four  "  etc. 


8  ELEMENTARY  ALGEBRA  AND  §  3 

The  words  major  and  minor  are  used  only  when  capitals 
and  small  letters  are  employed  in  the  same  problem.  Other- 
wise they  are  dropped,  and  a',  #2,  for  example,  are  read 
"  a  prime,  b  sub-two. " 


POSITIVE    AND    NEGATIVE    QUANTITIES. 

29.  Positive  and  negative  are  terms  applied  to  quan- 
tities of  opposite  character ;    as,  money  earned  and  money 
owed,  water  running  into  a  tank  and  water  running  out,  a 
distance  up  stream  and  a  distance  down  stream,  the  height 
of  a  tower  and  the  depth  of  a  well,  the  pull  on  a  lifting  rope 
and  the  weight  of  the  load,  etc. 

30.  Positive  quantities  are  preceded  by  the  sign  plus, 
as  -|-  %xy,  -{-  ab,  etc. ,  and   negative   quantities  by  the  sign 
minus,    as   —  %xy,  —  ab,    etc.       Thus,    if   money   earned   is 
-\-  $50,  a  like  amount  owed  is  —  $50.     If  the  quantity  of  water 
running  into  a  tank  is  denoted  by  -f-  #,  the  same  quantity 
running  out  should  be  denoted  by  —  a. 

31.  It  really  does  not  matter  which  quantity  be  taken  as 
positive  and  which  as  negative,  so  long  as  the  characters  of 
the  positive  and  the  negative  quantity  are  opposite ;  but  it  is 
customary  to  call  something  gained  positive  and  something 
lost  negative.     Thus,  money  earned  is  usually  regarded  as 
positive,    money  owed  as  negative;   distance  up,   positive, 
distance  down,  negative. 

32.  The  signs  -f  and  —  may  be  used  in  two  entirely  dif- 
ferent senses;  heretofore  we  have  used  them  exclusively  as 
symbols  of  operation ;  thus,  -|-  placed  between  two  quantities 
indicates  that  they  are  to  be  added,  etc.     In  the  distinction 
between   positive    and    negative    quantities,    however,    we 
denote  the  positive  quantity  by  the  sign  -}-  and  the  negative 
quantity  by  the  sign  — .     Hence,  under   different    circum- 
stances, these  signs  may  denote  addition  and  subtraction,  or 
they  may  denote  positive  and  negative  quantities.     Suppose 
we  write  the  expression  $500  —  $200  —  $300 ;  the  sign  —  in 


§  3  TRIGONOMETRIC  FUNCTIONS.  9 

this  case  indicates  that  the  $200  is  subtracted  from  the  $500. 
Suppose,  however,  that  a  man  has  in  his  possession  $500  and 
owes  $200.  The  former  amount  we  may  denote  by  -j-  $500, 
and  the  latter,  since  it  is  owed,  by  —  $200 ;  in  this  case  the 
sign  —  before  the  $200  indicates  the  negative  character  of 
the  quantity.  To  find  how  much  the  man  is  worth  we  add 
the  two,  thus: 

-f  $500  -f-  (-  $200)  =  +$300. 

In  this  addition  the  second  -j-  sign  denotes  the  operation 
of  addition,  while  the  three  signs  immediately  before  the- 
quantities  denote  the  positive  ~or*  negative  character  of  the 
quantities. 

33.  It  is  usual  to  consider  a  quantity  as  increasing  in  a 
positive  direction ;  any  positive  quantity,  no  matter  how  small, 
is  always  considered  greater  than  any  negative  quantity,  no 
matter  how  large. 

The  value  of  a  negative  quantity  is  conceived  to  increase 
as  its  numerical  value  decreases.  A  man  who  is  $10  in  debt 
is  better  off  than  one  who  is  $50  in  debt;  and  the  man  who 
has  $5  in  the  bank  is  better  off  than  either.  Thus,  5  is 
greater  than  — 10,  and  — 10  is  greater  than  —  50. 

34.  When  writing  algebraic  expressions,   if  a  positive 
term  stands  alone,  or  if  the  first  term  of  an  expression  is 
positive,  the  plus  sign  is  omitted,  it  being  understood  that 
the  term  is  positive.     Thus,  3#  means  the  same  as  +3#, 
and  a  —  b  the  same  as  -f-  a  —  b.     The  minus  sign  must  never 
be  omitted.     Polynomials  are  usually  written  with    a  pos- 
itive  term   first,  and  monomials  with  the  letters  arranged 
alphabetically. 

EXAMPLES  FOR  PRACTICE. 

35.  Express  the  following  algebraically: 

1.  Three  x  square/  square,  minus  two  cd  into  a  +  b. 

Ans.  Zx^-Zcd^d  +  b}. 

2.  The  quantity  ;;/  square  plus  two  mn  plus  n  square  in  parenthesis, 
times  a  square  b  cube  c  fourth.  Ans.  ( 


10  ELEMENTARY  ALGEBRA  AND  §  3 

3.  A,  plus  the  square  root  of  D,  times  the  parenthesis  Jfplus  V. 

Ans.  A  +  i/V(X+  Y). 

4.  A,  plus  the  radical  D  times  the  parenthesis  A'  plus  Y. 


Ans.  A  +  tfD(X+  Y}. 

5.  Ten  x  plus_y,  minus  seven  times  the  quantity  .r  minus  the  frac- 
tion^ over  4  in  parenthesis,  plus  the  fraction  x  square  minus  y  square 

over  two  cd. 

Ans. 

When  a  =  6,  b  =  5,  and  c  =  4,  find  the  numerical  values  of: 

6.  a*  +  2ad  +  t>*.  Ans.  6'  +  2  X  6  X  5  +  5a  =  121. 

7.  2«a  +  Me  -  5.  Ans.  72  +  60-5  =  1  27. 

8.  2ac5  —  a\a  +  b).  Ans.   11,892. 

9.  abc*  +  ab*c  —  a*bc.  Ans.  360. 
When  x  —  8  and  j/  =  6,  what  do  the  following  equal  : 


10.     (.r  +y)  (.r  -y)  - .{/  ^-  ?     Ans.  (8  +  6)  (8  -  6)  - 1/  -5^  =  26. 


11.     \/(jc  +y*)  (,r2  +  y)  -  (.1  - 


ADDITION   AND    SUBTRACTION. 


PRELIMINARY  IDEAS. 


36.     Suppose  we  take  a  point,  as  A  on  the  line  shown  in 
Fig.    1,  and  lay  off  equal  distances  in  opposite  directions. 


T  M  T  f  f  T  M  ]  f  T  f  f  f  T   M 


FIG.  1. 


Now,  let  us  call  distances  to  the  right  -f-,  or  positive,  and 
distances  to  the  left  — ,  or  negative.  Let  us  also  call  a  move- 
ment to  the  right  positive,  and  one  to  the  left  negative. 
Suppose  the  positive  direction  east,  the  negative  west,  and 
the  distances  to  represent  miles. 


g  3  TRIGONOMETRIC  FUNCTIONS.  11 

Suppose  a  man  starts  from  A  and  walks  east  6  miles,  and 
after  a  pause  walks  3  miles  farther  east.  His  distance  from 
A  is  +  ()  +  (+  3)  =  +9  miles,  the  plus  sign  being  taken 
because  the  motion  is  in  the  positive  direction.  Suppose, 
however,  the  man  starts  from  A  and  walks  west  3  miles,  and 
after  a  pause  walks  5  miles  farther  west.  His  distance  from 
A  is  8  miles  west  of  A,  or  8  miles  in  a  negative  direction; 
that  is,  —  3  -|-  (—  5)  =  —  8.  As  a  third  case,  imagine  the 
man  to  walk  6  miles  east,  and  then  turn  around  and  walk  4 
miles  west.  Counting  4  west  from  6,  we  see  that  he  would 
still  be  2  miles  east  of  A  ,  or  2  miles  in  a  positive  direction  ; 
that  is,  +C  +  (—  4)  =  2.  If,  instead  of  walking  back  4 
miles,  he  had  walked  back  10  miles,  we  find  by  counting  10 
miles  west  from  6  that  he  would  have  been  4  miles  west  of 
A,  or  4  miles  in  a  negative  direction,  +G  +  (—  10)  =  —4. 
For  reference,  the  above  results  are  collected  : 


(  +  3)  =  +9 
_3  +  (-5)  =  -8 


-f-6-K-lO)  =   -4 

The  student  should  observe  carefully  that  in  each  of  these 
additions  the  signs  immediately  before  the  numbers  denote 
their  positive  or  negative  character,  while  the  -f-  sign  m 
front  of  the  parenthesis  denotes  the  operation  of  addition. 

37.  From  these  illustrations  we  have  the  following 
important  principle  :  If  all  the  terms  to  be  added  arc  positive, 
the  sum  is  positive  ;  if  all  are  negative,  the  sum  is  negative. 
If  one  term  is  positive  and  the  other  is  negative,  the  sum  has 
the  sign  of  the  numerically  greater.  If  there  are  several 
terms  to  be  added,  part  of  which  are  positive  and  part  nega- 
tive, the  sum  is  positive  if  the  sum  of  the  positive  terms  is 
numerically  greater  than  the  sum  of  the  negative  terms. 

When  the  terms  have  the  same  sign,  the  numerical  sum  is 
that  which  would  be  obtained  if  the  signs  were  disregarded. 
Thus,  in  the  first  case  above,  the  signs  of  0  and  3  are  the 
same,  therefore  the  sum  is  numerically  0  +  3  =  9  ;  likewise 


12  ELEMENTARY  ALGEBRA  AND  §  3 

in  the  second,  the  signs  of  the  3  and  the  5  are  alike  and  the 
sum  is  numerically  3  -|-  5  =  8.  When,  however,  one  term  is 
positive  and  the  other  is  negative,  the  sum  is  the  numerical 
difference  between  the  terms.  Thus,  in  the  third  case  above, 
the  signs  of  6  and  4  are  different,  and  the  sum,  2,  is  the 
numerical  difference;  likewise  in  the  fourth  case,  the  0  and 
the  10  have  different  signs,  and  the  sum,  4,  is  the  numerical 
difference. 

38.  To  add  like  terms  containing  letters,  we  simply  add 
the  coefficients,  having  regard  for  the  proper  signs,  and 
annex  the  literal  part.  Thus,  the  sum  of  §ax*y  and  ^ax^y  is 
9a#y-t  the  sum  of  %ab  and  —  \\ab  is—  §ab\  and  the  sum  of 
9»/a//  3/;/2#,  —  8;//2//,  and  2m*n  is  6;/r;/. 


ADDITION  OF  MONOMIALS. 

39.  Like  Quantities. — To  add  like  quantities  having 
the  same  sign : 

Rule  I. — Add t lie  coefficients,  give  the  sum  the  common  sign, 
and  annex  the  common  literal  part. 

To  add  like  quantities  having  different  signs : 

Rule  II. — Add  the  positive  and  the  negative  coefficients 
separately,  and  from  the  greater  sum  subtract  the  less.  Give 
the  remainder  the  sign  of  the  greater  sum,  and  annex  the 
common  literal  part. 

EXAMPLE. — Find  the  sum  of  —  %abxy,  —  abxy,  —  *&abxy,  and  —  §abxy. 

SOLUTION. — The  sum  of  the  coefficients  is  12  (remember  that  the 
coefficient  of  —  abxy  is  1),  and  the  common  sign  is  — .  The  common 
literal  part,  abxy,  annexed  to  these  gives  as  the'  result  —  VZab.iy. 
(Rule  I.) 

EXAMPLE. — Combine  ,ry2,  —  2.rj/2,  8_ry2,  and  —  4.ry2. 

SOLUTION. — The  sum  of  the  coefficients  of  the  positive  terms  is  9, 
and  of  the  negative  terms,  6.  Their  difference  is  3,  and  the  sign  of  the 
greater  sum  is  +.  The  common  literal  part,  xy*,  annexed  to  these 
gives  as  the  result  3^/2.  (Rule  II.) 


§  3  TRIGONOMETRIC  FUNCTIONS.  13 

SUBTRACTION  OF  MONOMIALS. 

40.  Referring1  again  to  Fig".  1,  suppose  two  men,  C  and 
D,  start  from  point  A  and  travel  eastward.      At  the  end 
of  a  certain  time,  C  has  walked  8  miles  and  D  has  walked 
5  miles.     Now,  the  distance  between  C  and  D  is  3  miles;  to 
pass  from  -j-  5  to  +  8,  we  must  walk  +  3  miles  east,  or  in  a 
positive  direction ;  or  -f-  8  —  ( +  5)  =  +  3-     Observe  that  the 
minus  sign  here  denotes  subtraction,  while  the  three  plus 
signs  denote  that  the  three  quantities  which  they  precede  are 
positive.     The  difference  between  5  and  8  means  how  far, 
and  in  what  direction  we  must  go  to  pass  from  5  to  8  or  from 
8  to  5.     If  we  pass  from  -f-  5  to  -f-  8,  we  move  in  a  positive 
direction,  and  we  say +5  from +8  is  equal  to  +  3.     Sup- 
pose, however,  we  pass  from  +  8  to  +  5 ;  we  move  westward, 
or  in  a  negative  direction,  a  distance  of  3  miles.     Hence,  we 
say  that  +8   from  +5  is  -3,  or  +5  -  (  +  8)  =  -3.     We 
shall  always  consider  the  point  we  pass  from,  as  the  subtra- 
hend, or  quantity  to  be  subtracted,  and  the  point  we  approach 
as  the  minuend,  or  quantity  we  subtract  from. 

Suppose  C  has  walked  7  miles  east  and  D  has  walked 
4  miles  west,  how  far  apart  are  they  ?  To  pass  from  D 
to  C  we  must  travel  11  miles  east,  or  in  a  positive  direc- 
tion. Therefore,  7  —  (  —  4)  =  +11.  To  pass  from  C  to  D, 
we  travel  11  miles  west,  or  in  a  negative  direction; 
-4-(  +  7)  =  -11. 

The  following  exercises  may  be  studied  in  connection 
with  Fig.  1: 

From  +  3  to  +  8  is  +  5,  or  +8- ( +  3)  =  +  5. 
From  +  10  to  +  6  is  —  4,  or  +  6  —  (  +  10)  =  —  4. 
From  —  5  to  +  4  is  +  9,  or  +  4  —  ( —  5)  =  +9. 
From  - 9  to  —2  =  +  7,  or  - 2 -  ( - 9)  =  +  7. 
From  —  3  to  —  7 .  =  —  4,  or  —  7  —  ( —  3)  =  —  4. 
From  +  2  to  - 4  =  -  6,  or  - 4-  (  +  2)  =  -  6. 

41.  In  every  case,  the  difference  is  what  must  be  added 
to  the  subtrahend  to  obtain  the  minuend.     Thus,  if  I  am 
3  miles  east  of  A,  Fig.  1,  how  far  must  I  go  to  be  8  miles 
east  of  A  ?    Evidently  5  miles,  since  5  miles  added  to  3  miles 
gives  8  miles.     Similarly,   the  difference  between  —  5  and 


14  ELEMENTARY  ALGEBRA  AND  §  3 

» 

-|-4  is  -f-  9,  since  -f~9  must  be  added  to—  5  to  obtain  -|-  4; 
that  is,  we  must  travel  9  miles  east  in  passing  from  —  5  to 


4:2.  We  have  seen  that  if  we  add  -f-  6  and  —  4  we  obtain 
+  2  as  the  sum.  (Art.  36.)  If  we  subtract  -+-  4  from  -f  6, 
the  difference  is  -f-  6  —  (  -f-  4)  =  +  2,  since  -f-  2  must  be  added 
to  +  4  to  make  -j-  G.  If  we  add  +  6  and  —  10,  the  sum  is 
—  4.  If  we  subtract  +10  from  -{•  6,  the  difference  is  —  4. 
Therefore,  it  appears  that  if  we  wish  to  subtract  one 
quantity  from  another,  we  obtain  the  same  result  if  we 
change  the  sign  of  the  quantity  to  be  subtracted  and  add  it 
to  the  other  quantity.  Thus,  +  7  subtracted  from  -|-  12  is 
the  same  as  —  7  added  to  +  12,  the  result  being  5  in  either 
case.  —  3mn  subtracted  from  kmn  gives  the  same  result  as 
-f-  Sinn  added  to  4;;z;z.  The  following  exercises  are  given  as 
illustrations: 


=  +7. 
=  -6  +  (-3)  =  -9. 

+  2a  —  (  +  a)  =  +  2a  +  (  —  a)  =  +  a. 

+  §xy  —  (  +  IS.ry)  =  +  §xy  +  (  —  \§xy}  —  —  \§xy. 

43.  To  subtract  like  quantities  : 

Rule.  —  Change  the  sign  of  the  subtrahend,  and  proceed  as 
in  addition. 

EXAMPLE.  —  From  —  ZaPx  take  lab**. 

SOLUTION.  —  Changing  the  sign  of  the  subtrahend,  lab*x,  and  add- 
ing, we  have  —  ^ab^x  +  (  —  lab**)  =  —  IQab^x.  Ans. 

44.  Unlike  Quantities.  —  In   arithmetic,    unlike   num- 
bers,  as  5  books  and  3  dollars,  cannot    be  added  or  sub- 
tracted.   So,  in  algebra,  unlike  terms,  as  3##2,  ±xy,  2m,  etc., 
cannot  be  combined  or  subtracted,  except  by  indicating  the 
operations  by  signs. 

Expressions  in  algebra  are  composed  of  quantities  between 
which  operations  of  addition,  multiplication,  etc.  are  indi- 
cated. The  trinomial  mz  —  %mn  -\-  n*,  for  example,  is  the 
indicated  sum  of  ;;z2,  —  %mn,  and  «a,  and  it  is  to  be  considered 


§  3  TRIGONOMETRIC  FUNCTIONS.  15 

as  one  quantity,  in  the  same  way  that  an  arithmetical  sum, 
obtained  by  actually  performing  the  addition,  is  considered. 

EXAMPLE. — What  does  led*  —  %cx  —  cd*  +  §adx  +  2cd*  equal  ? 

SOLUTION. — In  this  case,  part  of  the  terms  are  like  and  part  unlike. 
Combining  like  terms,  led*  +  2ed*  —  ed3  =  8ed'\  Connecting  the  unlike 
terms  with  this  result  by  their  respective  signs,  we  have  as  the  final 
result  Zed*  —  Sex  +  Qadx.  Ans. 

EXAMPLE. — Subtract  2m  —  3  from  1m  +  %xy. 

SOLUTION. — As  in  the  example  above,  we  have  like  and  unlike  terms. 
Subtracting  like  terms,  Art.  43,  we  have  1m  —  2m  =  5m.  We  must 
now  connect  the  unlike  terms  by  their  respective  signs.  Since  —  3  is  in 
the  subtrahend,  its  sign  will  be  changed,  giving  us  5m  +  Zxy  -h  3.  Ans. 


EXAMPLES  FOR  PRACTICE. 

45.      Find  the  sum  of  the  following. 

1.  -  60 2,  2a\  -  50 2,  4# 2,  -  3« 2,  and  a\  Ans.  - 

2.  2a*t>,  -a*&,  lla*&,  -5a*t>,  4a26,  and  -  9«2£.  Ans. 

3.  2x*,  %xy,  —  x'\  8y2,  —  5xy,  and  —  7j/2.  Ans.  x*  —  Zxy  +y'\ 
NOTE. — Combine  like  terms  and  connect  with  respective  signs. 

4.  a>bc,  -2#£V,  Zabc-,  -IcPbc,  and  5a£V. 

Ans.  %ab *c  —  ZcPbc  •+-  §abc*. 

Solve  the  following : 

5.  From  \la  take  —  \\a.  Ans.  2Sa, 

6.  From  —  \\a  take  Via.  Ans.  —  2Sa. 
1.     Subtract  5cd  from  —  ±cd.  Ans.  —  $cd. 

8.  Subtract  - 10£2  from  -  10^2.  Ans.  0. 

9.  What  quantity  added  to  \§xy  will  produce  —  \%xy  ?  Ans.  —  22.ry. 
10.     What,  then,  does  IQxy  subtracted  from  —  \2xy  equal  ? 

Ans.  —  2%xy. 


ADDITION  AND  SUBTRACTION  OF  POLYNOMIALS. 

46.     To  add  polynomials: 

Rule. —  Write  the  expressions  underneath  one  another,  with 
like  terms  in  the  same  vertical  column.  Add  each  column 
separately,  and  connect  the  sums  by  their  proper  signs. 


1C  ELEMENTARY  ALGEBRA  AND  §  3 

EXAMPLE.—  Find   the  sum   of   5#2  +  Gac  —  3^2  —  2xy,  7  ac  —  3a*  +  4l>'2 
+  3,ry,  and  4.tj  —  5£2  +  Sac  —  a'2. 

SOLUTION.  —  Writing  like  terms  in  the  same  vertical  column,  we  have 


-   a'2  +   Sac  - 


sum      a2  +  21ac  —  4^-  +  5.ij.     Ans. 

EXAMPLE.  —  Find    the    sum    of    a?x  —  ax*  —  X*,    ax  —  x^—a1,    —  2a* 
-  2a'2.r  -  2ax\  and  3rt2  -  3a\r  +  %ax\ 

SOLUTION.  —  a^x—   ax"*  —   x* 


sum    —  ±a?x  +  0       —  2.ra  +  0     +  ax 

-  ax-±a*x-2x\     Ans.  (Arts.  23  and  34.) 

47.     To  subtract  one  polynomial  from  another: 

Rule.  —  Write  tJic  subtrahend  underneath  the  minuend, 
with  like  terms  in  the  same  vertical  column.  Change  the 
sign  of  each  term  of  the  subtrahend,  and  proceed  as  in  addi- 
tion. 

EXAMPLE.  —  From  Sac  —  2b  subtract  ac  —  b  —  d. 

SOLUTION.  —        Sac  —  Zb 

—   ac  +   b  -\-d,  subtrahend  with  signs  changed. 


difference      2ac  —    b  +  d.     Ans. 

EXAMPLE. — From  2.t3  —  3.t2/  +  2,rya  subtract  .r3  —  aj2  +_y3. 

SOLUTION. —    2.r3  —  3.r2^/  +  2.i;/2 

-   .i3  +   xy*  —y*,  subtrahend  with  signs  changed. 

difference      ,r3 


EXAMPLES  FOR  PRACTICE. 

48.      Find  the  sum  of  the  following: 

1.     ax  +  2bx  +  ±by  —  %ay,  2ax  +  bx  +  2ay  —  by,  and  4ax  +  %by. 


Ans.  lax  +  %bx  +  §by  —  ay. 

2.     a  —  x  +  4y  —  3^  +  wt  z  +  3a  —  2.v  —y  —  w,  and  x  +y  +  z. 

Ans.  4#  — 


3  TRIGONOMETRIC  FUNCTIONS.  17 


3.  2a  —  30  +  4</,  30  —  3(t  +  4<r,  2d  —  3c~  +  4«,  and  2c  —  3d  +  40. 

Ans.  3a  +  3/5  +  3c  +  3d. 

4.  (j.v  —  3y  +  7w,  2;z  —  ,v  +_/,  2y  —  4.tr  —  5/«,  and  ///  +  //  —  y. 

Ans.  x  —y  +  oni  +  3n. 

Solve  the  following  : 

5.  From  la  +  50  —  3r  take  a  —  Ib  +  §c  —  4. 

Ans.  6rt  +  120  --  &:  +  4. 

6.  From  3;«  —  5«  +  r  —  2j  take  2r  +  3w  —  ;«  —  5s. 

Ans.  4/«  —  8«  —  r  +  3.v. 

7.  Subtract  2.r  —  2y  +  2  from  j  —  .r.  Ans.  3y  —  3.v  -  2. 

8.  Subtract  3.r3  +  4.v2_y  —  7.rj/2  -(-j/3  —  .ry3  from  5.r3  +  x*y  —  6.rya  +  y\ 


SYMBOLS    OF    AGGREGATION. 

49.  Parentheses,  brackets,   etc.   being   used   to   enclose 
expressions  that  are  to  be  treated  as  one  quantity,  the  sign 
before  the  symbol  affects  the  entire  expression,  not  the  first 
term   only.      Thus,     —  (a?  —  %ab  -\-  ^2)  signifies  that  all  the 
terms  are  to  be  subtracted  from  what  precedes,  not  cf  only. 

50.  When  combining  the  terms  of  any  expression  with- 
out parenthesis,  we  proceed  as  in  addition  of  monomials. 
When  we  have  a  parenthesis  preceded  by  a  minus  sign,  we 
must  consider  the  expression  within   the  parenthesis  as  a 
subtrahend,   and   change   all    signs    before    removing    the 
parenthesis. 

If,  on  the  contrary,  the  sign  of  the  parenthesis  is  plus, 
we  may  remove  the  symbols,  but  we  must  not  change  the 
signs  of  the  expression,  because  we  do  not  change  the  signs 
of  an  expression  to  be  added. 

51.  When  a  quantity  is  enclosed  by  a  parenthesis,  the 
first  term  is  understood   to  have  the  plus  sign,  unless  the 
minus  sign  is  given;  thus,  in  the  expression  —  (8.r  +  5  —  2£), 
the  minus  sign  refers  to  the  whole  quantity.     The  sign  of 
8^"  is  -}-,  and  the  expression  if  written  in   full  would  be 

5-**). 


18  ELEMENTARY  ALGEBRA  AND  §  3 

EXAMPLE.  —  Remove  the  parenthesis  from  4c  —  (8«  +  kab  —d}. 

SOLUTION.  —  Changing  the  sign  of  each  enclosed  term,  and  remem- 
bering that  the  sign  of  3a  is  +,  understood,  we  have  as  the  result 
4c  —  'da  —  4afr  +  d.  Ans. 


EXAMPLE.  —  Remove  the  parentheses  from  4a  —  5^  —  (a  —  4.*)  +  (x  —  Sa). 
SOLUTION.  —    4a  —  5x  —  (a  —  4r)  +  (x  —  8a)  =  4a  —  5x  —  a  +  4,r  +  x  —  8a. 
Adding  the  like  terms,  we  have 


=  —5a.     Ans. 

52.  Symbols  of  aggregation  will  often  be  found  enclo- 
sing others.  In  such  cases  they  may  be  removed  in  suc- 
cession, always  beginning  with  the  innermost  pair. 

EXAMPLE.  —  Remove  all  the  symbols  of  aggregation  from 

§a—\b  —  [led  -  4«  +  (Zed  -  a^6)~\  }  . 

SOLUTION.  —  We  first  remove  the  vinculum.     This  being  in  effect  the 
same  as  the  parenthesis,  the  minus  sign  before  the  a  indicates  that 
+  a  and  —  b  are  to  be  subtracted. 
Hence,  we  have 

Qa  _  j  ft  _  ^lcd-^a  +  (2cd-  a  +  b)}  }  . 
Removing  the  parenthesis  we  have 


This,  with  the  brackets  removed,  is  equal  to 

§a-\b 
which,  in  turn,  is  equal  to 


—a  +  b. 
Combining  like  terms, 

6«  —  4a  —  a  —  b  +  b  +  led  +  Zed  =  a  +  §cd,     Ans. 


EXAMPLES  FOR  PRACTICE. 

53.      Remove  the  parentheses  from  the  following: 

1.  —  (2mn  —  m"2  —  »2).  Ans.  ;«2  —  2mn  +  ri2. 

2.  \-(-b  +  c  +  '$).  Ans.  b  —  c  —  %. 

3.  5a-4t>  +  3c-(-3a  +  2t>-c).  Ans.  80-6£+4<r. 

4.  3.r  -  (2*  -  5)  +  (7  -  x).  Ans.  12. 
Remove  the  symbols  of  aggregation  from  the  following: 

5.  m-±n-k-m  +  n-Zk.  Ans.  2?n-'6n-k. 


TRIGONOMETRIC  FUNCTIONS.  19 


6.  5,r  —  (2x  —  3y)  —  (x  +  5y).  Ans.  2.r  — 

7.  3a-7tf-(5«-£-*)--tf-4£).  Ans. 


8.  3or  +  2y  —  [5.r  —  (3y  +  x  —  4y)]  .  Ans.  j  —  .r. 

9.  100,r-{200,r-[500.r-(-100ar)-300.r]-400^}.          Ans.  600,r. 
10.     lcx  —  ±c—tcx  +  Sc     +  c    —  cjc.  Ans. 


NOTE.  —  Observe  that  the  sign  before  the  parenthesis  is  +  under- 
stood. 


MULTIPLICATION. 


PRELIMINARY  IDEAS. 

54.  In  algebra,  multiplication  is  often  indicated  only, 
and  the  final  answer  is  frequently  nothing  more  than  the 
symbols  so  written  that  it  is  shown  they  should  be  multiplied 
when  their  numerical  values  are  substituted.  Thus,  the 
product  of  m  and  n  is  ;//;/,  the  absence  of  a  sign  between  the 
two  quantities  denoting  the  operation  of  multiplication.  If 
we  multiply  2a  by  3£,  the  four  factors  which  form  the  prod- 
uct are  2,  a,  3,  and  b\  hence,  Za  X  3£  =  2  X  a  X  3  X  b.  Since 
we  are  at  liberty  to  arrange  the  factors  in  any  order,  we  have 


Hence,  in  finding  the  product  of  two  quantities,  the  coeffi- 
cients are  multiplied  together  and  prefixed  to  the  literal 
factors. 

55.  When   two  quantities  are  multiplied  together,  the 
sign  of  the  product  is  positive  if  the  tivo  quantities  have  the 
same  sign;  and  it  is  negative  if  tJie  quantities  have  opposite 
signs. 

Thus,  (  +  5«)x(  +  3/«)  =  +15;«», 

(  —  5«)X(  +  3*«)  =  —  Wmn. 
(  —  5«)X(  —  3/«)  =  +15mn. 
(  +  5»)X(  —  3*»)  =  —  15mn. 

56.  Assume  that  the  3;«  above  is  the  multiplier,  and 
that  5//  is  the  multiplicand.     The  multiplier  must  be  con- 
sidered as  an  abstract  number,  and  simply  shows  how  many 


20  ELEMENTARY  ALGEBRA  AND  §  3 

times  the  multiplicand  is  to  be  taken  ;  thus,  3m  may  repre- 
sent 15,  21,  30,  etc.,  depending  upon  the  value  of  m.  The 
multiplicand,  5;/,  may  represent  any  concrete  number;  as, 
$20,  GO  feet,  100  miles.  The  sign  of  the  mtiltiplicand  shows 
the  character  of  the  quantity;  thus,  if  -f-5;z  represents  $20, 
—  5n  may  be  taken  to  represent  a  debt  of  $20.  The  sign  of 
the  multiplier,  on  the  other  hand,  is  a  symbol  of  operation, 
and  shows  how  the  result  is  to  be  treated  after  it  is  obtained. 
If  the  multiplier  is  positive,  the  result  obtained  is  to  be 
added  when  taken  in  connection  with  other  quantities;  the 
multiplier  is  given  the  negative  sign  to  show  that  the  result 
is  to  be  subtracted. 

Multiplying  +5/2  by  3/#,  considering  the  multiplier  sim- 
ply as  an  abstract  number,  and  disregarding  its  sign,  we 
obtain  -j-15;//;/.  Similarly,  —  5n  multiplied  by  3m  gives 
-  15  ;////.  The  product  in  each  case  has  the  same  char- 
acter as  the  multiplicand.  Now,  if  the  multiplier  has  the 
positive  sign,  the  product  is  to  be  added  when  taken  in 
connection  with  other  quantities,  and,  therefore,  the  signs 
remain  as  above.  That  is,  (  +  on)  X  (  +  3m)  —  -f-  15mn,  and 
(-50)  x"(  +  3;«)  =  —I5mn. 

If,  on  the  other  hand,  the  multiplier  has  the  negative 
sign,  the  product  is  to  be  subtracted,  and,  therefore,  its 
sign  must  be  changed  when  combined  with  other  quanti- 
ties. Thus,  when  the  multiplier,  3m,  has  the  negative  sign, 
we  have  (  +  5n)  X  (  —  3m)  =  —  I6mn  and  (  —  5n)  X  (  —  3  m) 


57.  Exponents.  —  It  has  been  shown  that  a*  =  ay.  a 
3  =  axaxa.  The  product  a*xa\  or  0V  =  aXaXaXaXa 
=  a*.  The  exponent  2  shows  that  a  is  used  twice  as  a  factor, 
and  the  exponent  3  shows  that  a  is  used  three  times  as  a 
factor.  In  the  product,  a  must  be  used  five  times  as  a  factor, 
or  the  exponent  of  #is  5.  The  exponent  of  the  product  is  the 
sum  of  the  exponents  of  the  factors. 

Thus,  rt4X«3  =  a*+9  =•  a\ 


§  3  TRIGONOMETRIC  FUNCTIONS.  21 

MULTIPLICATION  OF  MONOMIALS. 

58.  Rule.  —  To  the  product  of  tJic  coefficients  annex  the 
letters  of  bot  h  factors;  give  each  letter  an  exponent  equal  to 
t  he  sum  of  t  lie  exponents  of  tliat  letter. 

Make  the  sign  of  the  product  plus,  when  the  signs  of  the 
factors  are  alike;  and  minus,  when  they  are  unlike. 

EXAMPLE.  —  Multiply  kofb  by  —§a*bc. 

SOLUTION.  —  The  product  of  the  coefficients  is  20,  and  the  letters  to 
be  annexed  are  a,  b,  and  c.  The  new  exponent  of  a  is  5,  and  of  b,  2, 
since  a*+s  —  a5,  and  ^1+1  =  b^.  The  sign  of  the  product  is  minus, 
since  the  two  factors  have  different  signs.  Hence,  4a'2&  X  —  §<iAbc 
=  -20«5£V.  Ans. 

59.  When  there  are  more    than  two   factors,   we  have 
simply  three  or  more  examples  in  multiplication  to  solve  in 
succession,  each  to  be  performed  by  the  foregoing-  rule. 

EXAMPLE.  —  Find    the  continued  product  of  G.r2^3,   —  9.r2_y25'2,  and 


SOLUTION.—    6.i-2^3  X-9.rj2^2  =  -54.r2+2j/'+2.23+2,   or  - 
Now,    multiplying      this     product    by  —  S.^ys,    we    have  — 
6.     Ans. 


EXAMPLES  FOll  PRACTICE. 

6O.      Find  the  product  of: 

1.  rt3£2  and -5rt&/.  .  Ans.  -5a*6*it. 

2.  —  Ixy  and  —  7.r2j/2.  Ans.  49.v3j/3. 

3.  —  15w5«6  and  3mn.  Ans.  —45wGn\ 

4.  3<z(jr  —  j/)2  and  2a\x  —  y).  Ans.  6rt3(.r— j)3. 
SUGGESTION. — Treat  the  (x—y)  as  though  it  were  a  single  letter. 

5.  Find  the  continued  product  of  2a^m<2.v,  —  3rt2;/z.t3,  and 

Ans.  —  2 

6.  What  does  —  a^bn  X  —  %cdn  X  —  §bdc*  X  —  2acn'2  equal  ? 

Ans.  12tf3 


MULTIPLICATION  OF  POLYNOMIALS. 

61.     When  one  of  the  factors  is  a  monomial : 
Rule. — Multiply  each  term  of  t  lie  polynomial  by  tlie  mono 
mial,  and  connect  the  separate  products  by  their  proper  signs. 


22  ELEMENTARY  ALGEBRA  AND  §  3 


EXAMPLE.—  Find  the  product  of  —  9«5  +  3«3^2  —  402£3  —  1>*  and  — 
SOLUTION.—  -  9«5  +  3a*&'*  -  4«2£3  -  bb 


.     Ans. 

62.     When  both  factors  are  polynomials: 
Rule.  —  Multiply  each  term  of  one  polynomial  by  each  term 
of  the  other,  and  add  the  partial  products. 

EXAMPLE.  —  Multiply  Qa  —  4£  by  4«  —  2£. 

SOLUTION.  —  Write  the  multiplier  under  the  multiplicand,  and  begin 
to  multiply  at  the  left  instead  of  at  the  right,  as  in  arithmetic,  since 
polynomials  are  always  written  and  read  from  the  left,  and  there  are 
no  numbers  to  carry. 

6«  -  4£  (1) 

4a  -  26 

Multiplying  (1)  by  4rt  gives      24a*  —  I6a&  (2) 

Multiplying  (1)  by  -  2£  gives  _      -12d£  +  8£2     (3) 

Adding  (2)  and  (3)  gives  2402  -  28a&  +  8£2.     Ans. 

It  will  be  noticed  that  the  like  terms,  —\§ab  and  —I2a&,  are  written 
under  each  other,  so  that  it  will  be  easier  to  add  them. 

EXAMPLE.  —  Multiply  x*  —  x  +  1  +  x'*  by  1  —  x1  +  x. 

SOLUTION.  —  With  a  view  to  bringing  like  terms  in  the  same  columns, 
arrange  both  multiplicand  and  multiplier  either  according  to  the 
increasing  or  the  decreasing  powers  of  the  same  letter.  (Art.  23.) 
Arranging  in  this  case  according  to  the  increasing  powers  of  x,  we 
have 

l-x  +  x2  +  x*  (1) 


Multiplying  (1)  by  1  gives  1  —  x  +  x*  +  x3  (2) 

Multiplying  (1)  by  +  ogives  x  —  x*  +   x*  +  x4  (3) 

Multiplying  (1)  by  —  x*  gives  —  ,r2+   x*  —  x*  —  x'a  (4) 

Adding  (2),  (3),  and  (4)  gives  1        -.r2  +  3^r3          -x\  Ans. 


63.  Multiplication  is  frequently  indicated  by  enclosing 
each  of  the  quantities  to  be  multiplied  in  a  parenthesis. 
The  sign  of  multiplication  is  not  placed  between  the  paren- 
theses, multiplication  being  understood.  When  the  quanti- 
ties are  multiplied  together,  the  expression  is  said  to  be 
expanded. 

For   example,   in    the   expression    (in  —  2n)  (2  ;#  —  //),    the 


§  3  TRIGONOMETRIC  FUNCTIONS.  23 

binomial  m  —  2n  is  to  be  multiplied  by  the  binomial  2m  —  n. 
Performing  the  multiplication,  the  product  is  2;/z2  —  5mn 
which  is  the  expanded  form  of  the  expression. 


EXAMPLES  FOB  PRACTICE. 

64.      Multiply  the  following  : 

1.  x*  +  2xy  +y*  by  x  +y.  Ans.  .t3 

2.  3«£a/«8  +  40V  -  2  by  aWm*.    Ans. 

3.  ^  _  ^    by  ^  +  ^ 

4.  .r4  +  .r2j2  +j/4  by  ^r2  —  j/2.  Ans.  ^-6  —  j6. 

5.  3«2-7rt  +  4by  2«2  +  9«-5.  Ans.  6«4  +  13^3  -  70«2  +  lla  -20. 
Expand  the  following  : 

6.  (2a  —  &)  (4  —  3«).  Ans.  8«  —  12^r  —  6«2  +  9^. 

7.  (.r  +  2)  (JT  -  2)  (-r2  +  4).  Ans.  .r4  ~  16. 

8.  .r^2-2 


NOTE.  —  The  expressions  in  the  brackets  reduce  to  x*  —  xy*  —  2  and 
^3  +  Xy*  +  2.     The  product  of  these  is  x6  —  x^  —  ±xy*  —  4.     Ans. 


THREE  IMPORTANT  EXAMPLES. 

65.  Let  a  and  b  be  any  two  quantities;  we  wish  to  find 
the  forms  of  the  following  products: 

(a  +  b}\  (a-b)\  and  (a  +  b)(a-b). 

By  actual  multiplication  we  find 

(a  +  by  =  c?  +  ^ab  +  b\  (1.) 

(a  -  by  =  a2  -  %ab  +  b\  (2.) 

(a  +  b)(a-b}  =  a*-b\  (3.) 

Hence  : 

66.  The  square  of  the  sum  of  two  quantities  is  equal  to 
the  square  of  the  first,  plus  twice  the  product  of  the  first  and 
the  second,  plus  the  square  of  the  second. 

The  square  of  the  difference  of  two  quantities  is  equal  to  the 
square  of  the  first,  minus  twice  the  product  of  the  first  and 
the  second,  plus  the  square  of  the  second. 

The  product  of  the  sum  and  difference  of  tivo  quantities  is 
equal  to  the  difference  of  their  squares. 


24  ELEMENTARY  ALGEBRA  AND  §  3 

67.  The  foregoing  statements  should  be  committed  to 
memory,  since  their  use  will  frequently  save  tedious  calcula- 
tion. The  student  is  advised  to  practice  until  he  is  certain 
that  he  knows  them  perfectly. 

EXAMPLE.  —  Square  3a'2  +  5. 

% 
SOLUTION.  —  The  square  of  the  first  term  is  3.t2  X  3.r2  =  9.r4  ;  twice 

the  product  of  the  terms  is  30.*"  ;  and  the  square  of  the  last  term  is  25. 
Hence,  by  formula  1,  letting  a  =  3.t2  and  b  =  5, 

(3.ra  +  5)2  =  9x*  +  30.r2  +  25.     Ans. 

EXAMPLE.  —  Square  fad—  x. 

SOLUTION.  —  The  square  of  the  first  term  is  IG^:2^2;  twice  the  product 
of  the  first  and  the  second  is  —  8cd.r  ;  and  the  square  of  the  last  term 
is  .r2.     Hence,  by  formula  2,  letting  a  =  4^/and  b  •=  x, 
(fad—xf  -  IGcW  —  Scdx  +  x*.     Ans. 


EXAMPLE.—  Expand  (.r2  +  3)  (.r2  -  3).     (See  Art.  63.) 

SOLUTION.  —  The  square  of  the  first  term  is  .v*,  and  of  the  second,  9. 
Hence,  by  formula  3,  letting  a  =  x*  and  b  =  3, 

(.t2  +  3)  (,r2  —  3)  =  .r4  —  9.     Ans. 


EXAMPLES  FOR  PRACTICE. 

68.      Square  the  following  : 

1.  m  +  n.  Ans. 

2.  4.1- +  2.  Ans.   IG.t-  +  16.V  +  4. 

3.  3«-5£.  Ans.  9#2  -  30^  +  25£2. 

Expand  the  following  : 

4.  (m  +  1)  (m  —  1).  Ans.  ;«a  — 1. 

5.  (x*+y*)(jc*-y*).  Ans.  x*-y*. 

6.  (4rt  +  4£2)(4tf-4£2).  Ans.  16«2  -  16£4. 

7.  Square  2c*  —  c  +  d. 

NOTE. — First  separate  2^2  —  c  +  d  into  two  terms  by  enclosing  c  +  d 
in  parenthesis ;  then  the  expression  becomes  2^2  —  (c  —  d*),  and  consid- 
ering this  as  a  binomial  we  find  the  square  to  be  4c*  —  fa"2(c  —  d) 
+  (c-d)\ 

-fa'\c-d}  =  _4^ 
(c-df  =  c*-2c 

Adding  these  results   to  4<r4,  the   final  result  is  4c4  —  4c* 
Ans. 


TRIGONOMETRIC  FUNCTIONS.  25 

DIVISION. 


INTRODUCTORY. 

69.  When  two  quantities  are  given,  and  we  wish  to  find 
a  third  quantity  which,  if  multiplied  by  one  of  the  first  two, 
will  produce  the  other,   the  process    of    finding  this  third 
quantity  is  called  division.    Thus,  if  the  given  quantities  are 
ab  and  a,  and  we  wish  to  find  a  quantity  which,  if  multiplied 
by  a,  will  give  ab,  we  must  divide  ab  by  a ;  our  quotient  will 
be  b,  since  a  x  b  =  ab.     Division  is,  therefore,  the  inverse  of 
multiplication. 

70.  The  following  laws  of  division  follow  directly  from 
the  statements  of  Arts.  54  to  5  7: 

If  the  dividend  and  the  divisor  have  like  signs,  the  quotient 
will  liavc  the  plus  sign ;  if  they  have  unlike  signs,  the  quo- 
tient will  have  the  minus  sign. 

The  coefficient  of  the  quotient  is  equal  to  the  coefficient  of 
the  dividend  divided  by  the  coefficient  of  the  divisor. 

The  exponent  of  a  letter  in  the  quotient  is  equal  to  its 
exponent  in  the  dividend  minus  its  exponent  in  the  divisor. 

71.  Let  it  be  required  to  divide  a*  by  a'\     We  have  to 
obtain  a  quotient,  which,  when  multiplied  by  the  divisor  a*, 
will  produce  the  dividend  a*.     The  quotient  is  evidently  1. 
By  Art.   7O,  however,  we  know  that  a^-t-a*  =  <72~2  =  a°. 
Hence,  any  quantity  whose  exponent  is  0  is  equal  to  1. 

From  the  foregoing  principles,  the  rules  for  division  are 
obtained. 

DIVISION  OF  MONOMIALS. 

72.  Rule. — Divide  the  coefficient  of  the  dividend  by  the 
coefficient  of  the  divisor  and  to  the  quotient  annex  the  letters 
of  the  dividend,  each  with  an  exponent  equal  to  its  exponent 
in  the  dividend  minus  its  exponent  in  the  divisor,  omitting 
those  letters  whose  exponents  become  zero. 


26  ELEMENTARY  ALGEBRA  AND  §  3 

Make  the  sign  of  the  quotient  plus  when  the  dividend  and 
divisor  have  like  signs,  and  minus  when  they  have  unlike 
signs. 

EXAMPLE.— Divide  6#50V3  by  —  3«afc8. 

SOLUTION. — The  quotient  of  6 -=-3  is  2.  The  letters  to  be  annexed, 
and  their  exponents,  are  «5~2  =  a*,  and  04~*  =  b*.  The  c  has  an 
exponent  of  3  — 3  =  0,  so  that  it  becomes  equal  to  1,  and  is  omitted. 
The  sign  of  the  quotient  is  minus.  Hence,  6a5& V3  -= —  §a*bc*  =  —  2a*fr*. 

Ans. 

PROOF.—    -3«2&:3X-2tf303  =  6*6£V3. 

EXAMPLE. — Divide  —  10#80V2*f  by  —  %abzc. 

SOLUTION.—    — 10«fl0W-f--200V  =  5aR-1^-3^-1^  =  §a*cd.     Ans. 


EXAMPLES  FOR  PRACTICE. 

73.      Divide  the  following  : 

1.  12wa#  by  4».  Ans. 

2.  30^5^3by-6,r5jV2.  Ans.  - 

3.  _44«3^V3by-ll^V3.  Ans. 

4.  -  lOO^3-^2  by  ^!r2-  Ans.  - 

5.  75/^2^3^4  by  75^3.  Ans. 


DIVISION  OF  POLYNOMIALS. 

74.     When  the  divisor  is  a  monomial: 

Rule.  —  Divide  each   term  of  the  dividend  by  the  divisor, 
and  connect  the  partial  quotients  by  their  proper  signs. 

EXAMPLE.—  Divide  12tf2£4-9fl£3  +  6tf3£4  by  300s. 
SOLUTION.—  300')  12a2£4-9d 


quotient      ±ab    —3       +2#20.     Ans. 


EXAMPLES  FOR  PRACTICE. 

75.      Divide  the  following: 

1.  64w2^3  —  32w»2  +  8;«2#by  Smn.  Ans. 

2.  27^3>/22'  —  9^3j/^2  —  333^3^2^2  by  —  3^3j2-.    Ans.  — 

3.  iO(jr + yY  -  5a(*  +y) + 5« 2(*  +y)  by  5(-^  +^)- 

Ans. 


§  3  TRIGONOMETRIC  FUNCTIONS.  27 

76.     When  the  divisor  is  a  polynomial: 

Rule. — Arrange  both  dividend  and  divisor  according  to 
the  ascending  or  descending  powers  of  some  letter. 

Divide  the  first  term  of  the  dividend  by  the  first  term  of 
the  divisor  for  the  first  term  of  the  quotient. 

Subtract  from  the  dividend  the  product  of  the  divisor  and 
this  term  of  the  quotient. 

Treat  the  remainder  as  a  new  dividend,  and  proceed  as 
before,  until  there  is  no  remainder,  or  until  the  final  remain- 
der contains  no  term  which  is  divisible  by  the  first  term  of 
the  divisor. 

EXAMPLE.  —Divide  x*  +  x*  —  9-r2  —  16.r  —  4  by  x*  +  4.r  +  4. 

SOLUTION. — 

quotient. 
+   ^3_   9^»_i6jr_4j:a-3.r-l.     Ans. 


_     x>-  4.r-4 

0         00 

The  first  term  x*  of  the  divisor  is  contained  in  x*,  the  first 
term  of  the  dividend,  x*  times;  hence,  x*  is  the  first  term 
of  the  quotient.  The  whole  divisor  multiplied  by  this  term 
gives  x*  -f-  4;tr3  +  4^r2  as  a  product,  which  subtracted  from  the 
dividend  gives  as  a  remainder,  —  3x3  —  13x*  —  16^  —  4.  It  is 
not  necessary  here  to  bring  down  the  —  4,  since  only  three 
terms  are  required  to  contain  the  divisor. 

The  first  term  x*  of  the  divisor  is  contained  in  —  3;r3, 
the  first  term  of  the  new  dividend,  —  3x  times.  Multiplying 
the  divisor  by  this  new  term  of  the  quotient,  we  have 
—  3x3—  ISx*  —  \%x.  Subtracting  this  from  the  first  remainder, 
we  obtain  —  x*  —  kx  —  4  for  a  new  remainder,  the  —  4  being 
brought  down  from  the  original  dividend.  The  first  term  of 
the  divisor  is  contained  in  the  first  term  of  the  new  remain- 
der or  dividend,  —  1  times.  Multiplying  the  divisor  by  this, 
we  get  —  x*  —  ±x  —  4,  which  subtracted  from  —  x*  —  ±x  —  4, 
the  last  remainder,  leaves  a  difference  of  zero.  The  work 


28  ELEMENTARY  ALGEBRA  AND  §  3 

ends  here,  since  there  are  no  more  terms  in  the  dividend  to 
be  brought  down. 

ExAMPLE.-^Divide  9.r2j2  +  .r4  -  4y 4  -  Gx3y  by  ^  +  2y*  -  Zxy. 

SOLUTION. — First  arrange  the  dividend  and  divisor  according  to  the 
descending  powers  of  x. 

*  —  4j/4  ( .r2  —  3.ry  —  2y2.    Ans. 


EXAMPLES  FOR  PRACTICE. 

77.      Divide  the  following : 

1.  ,r«  _  ix  + 12  by  x  —  3.  Ans.  .r  —  4. 

2.  .r2  +  x  -  72  by  .r  +  9.  Ans.  x  -  8. 

3.  2.v3  —  .r2  +  3.v  —  9  by  2.r  —  3.  Ans.  a-2  +  ,r  +  3. 

4.  .v4  +  II*8  -  12.v  -  5.r3  +  6  by  3  +  .r2  -  3.r.  Ans.   .ra  -  2.r  +  2. 

5.  ,r4  —  6.17  —  9,i-2  —  y1  by  ,ra  +_K  +  3.i .  Ans.  .r'2  —  3.r  —  y. 

6.  *"  —  1  by  x  —  1.  Ans.  *s  +  .r4  +  .r3  +  .i-'2  +  .v  +  1. 


FACTORING. 


78.  Factoring  is  the  process  of  finding  the  factors  of  a 
quantity,  that  is,  the  quantities  or  numbers  which  will  .divide 
that  quantity  without  a  remainder. 


79.  Expanding-  603£(2#-frt),  we  have 
hence,  Mb  and  (2#  +  tf)  are  the  factors  of  l^Y?2  -f  6«  3£. 
The  monomial  factor  Mb  may  be  further  resolved  into 
.3  X  2  X  tf  X  a  X  #.  In  solving  examples  in  factoring-,  it  is  not 
customary  to  write  out  the  factors  of  a  monomial,  since  they 
are  generally  apparent. 

That  the  student  may  be  able  to  recognize  factors  without 
the  labor  of  actual  division,  several  methods  of  readily 
discovering  factors  are  here  given. 


§  3  TRIGONOMETRIC  FUNCTIONS.  29 

80.  Equal    factors   are    those   whose    terms   have   the 
same  letters,    and  whose  letters  have  the  same  exponents 
and    the  same  signs.     Thus,   5a(%y  —  x)  and  5a(2y  —  x)  are 
equal  factors  of  5#(2j  —  x)  X  5  a  (by  -  .v)  =  25a*(2y  -  x}* ;  but 
5#(2j  —  ;r)  and  —  5a(2y  —  x)   are    unequal  factors,    since    the 
signs  of  5a  are  not  the  same  in  both  expressions. 

81.  A  product  of  two  equal  factors  is  a  perfect  square. 

Either  of  the  equal  factors  of  a  quantity  is  called  its  square 
root. 

82.  A  product  of  three  equal  factors  is  a  perfect  cube. 

Any  one  of  the  equal  factors  of  a  quantity  is  called  its  cube 
root. 

83.  In  factoring,  it  is  important  to  be  able  to  easily  dis- 
,tinguish  quantities  that  are  perfect  squares  and  cubes,  and 
to  determine  their  roots.      By  definition,  9#2$2  is  a  perfect 
square,  because  3ab  X  %ab  =  9#  2#2,  and  %ab  is  its  square  root. 
Also,  8ae>  is  a  perfect  cube  because  2<?2  X  %a*  X  2^2  —  8ar\  and 
2#2  is  its  cube  root.     In  each  of  these  cases  the  coefficients 
of  the  roots  are    multiplied   together,    and   the    exponents 
added,   to  produce  a  perfect  power.      Hence,  a  quantity  is 
a  perfect  square  when  its  coefficient  is  a  perfect  square,  and 
the  exponents  -of  all  its  letters  can  be  divided  by  2.       For 
example,  36;tr10,  49#W8,  16#6$12,  and  1  are  all  perfect  squares 
whose  roots  are  6^5,   lbc*d*,  4«3^8,  and  1,  respectively.     No 
perfect  square,   however,   can  have  a  minus  sign;    for,   let 
a  —  any   quantity,    —aY^  —  a  —  a*,    and   aXa  =  a*.      The 
square  root  of  «2  may  be  —  «,  or  a,  and  a  square  root  is  often 
written  ±  a,  read  plus  or  minus  a. 

A  quantity  is  a  perfect  cube  when  its  coefficient  is  a  perfect 
cube,  and  the  exponents  of  all  its  letters  can  be  divided  by  3. 
Thus,  27-r12,  —  64£VV,  8#I5£18,  and  1  are  all  perfect  cubes 
whose  roots  are  3x\  —±bc*d*,  %a'b\  and  1,  respectively. 
The  sign  of  the  cube  root  is  always  the  same  as  that  of 
its  cube 


30  ELEMENTARY  ALGEBRA  AND  §  3 

CASE  I. 

84.  When  all  the  terms  of  an  expression  are  divisible  by 
the  same  quantity,  the  expression  may  be  resolved  into  two 
factors  by  dividing  it  by  that  quantity. 

EXAMPLE.—  Factor  8m*n*  —  !Qm3ny  +  2m*n'2. 

SOLUTION.  —  We  first  examine  the  polynomial  to  see  if  one  of  its  terms 
is  contained  in  each  of  the  others.  Beginning  with  the  smallest  coeffi- 
cient, 2,  we  find  it  to  be  contained  in  each  of  the  others.  We  next  take 
the  literal  portion  m'*n2  of  this  term,  and  find  both  of  its  letters  in  each 
of  the  other  terms,  with  an  equal  or  higher  exponent.  Dividing  through 
by  2#z2«2,  we  have  for  our  quotient  4;/?2;z  —  §my  +  1.  Hence,  we  have 

1).     Ans. 


'EXAMPLE.—  Ascertain  if  12a£V3  -  18«V2j  +  24«Y4  -  36rt2^-5/2  has  a 
monomial  factor. 

SOLUTION.  —  By  inspection  we  find  that  the  smallest  coefficient  is  not 
contained  in  the  other  three  without  a  remainder.  This  coefficient,  12, 
is  the  product  of  3  X  2  X  2.  Of  these,  both  2  and  3  will  divide  all  the 
coefficients  once.  Therefore,  2x3  =  6  is  the  largest  numerical  factor. 
The  letters  a  and  c  are  contained  in  all  the  terms  ;  the  smallest  expo- 
nent of  a  is  1  and  of  c  is  2.  The  monomial  factor  is  evidently  6«^2. 
Dividing  the  polynomial  by  6tf^2,  the  quotient  is  2^  V  —  Sa^y  +  4ac'2 
the  factors  are  6ac*  and  2£  V  —  3«2  +  to2  —  §abc**.  Ans. 


EXAMPLES  FOR  PRACTICE. 

85.      Factor  the  following  expressions: 

1.  a*  +  ax.  Ans.  a(a3  +  x). 

2.  12rt5-2rt3  +  4rt4.  Ans.  2«3(6rt2  -1  +2a). 

3.  30w4«2  —  6#3.  Ans.  6«2(5w4  — «). 

4.  16.r>3-8^5  +  8.  Ans. 

5.  4x*jr  —  12*y*  +  S.rj/3.  Ans. 

6.  49«2£V4  -  63«3£V4  +  7rt4£V3.  Ans.  7« 


CASE  II. 

86.      To  factor  a  trinomial  which  is  a  perfect  square: 
Any  trinomial  is  a  perfect  square  when  the  first  and  the 
last  term  are  perfect  squares  and  positive,  and  the  second 
term  is  twice  the  product  of  their  square  roots. 


§  3  TRIGONOMETRIC  FUNCTIONS.  31 

Thus,  let  a  and  b  represent  any  two  quantities  whatever, 
and  we  have  the  general  forms  of  the  square  as  follows  : 

«2  +  2tf£  +  £2  =  (a+b)(a  +  b)  =  (a  +  b)\  (4.) 

a*  -  %a&  4-  {,*  =  (a  -  b)  (a  -  b)  =  (a-  b)\  (5.) 

These,  it  will  be  seen,  are  simply  the  inverse  of  formulas 
1  and  2,  Art.  65.  The  sign  of  the  second  term  of  the 
square  always  determines  the  sign  of  the  second  term  of  the 
root,  b  in  this  particular  case. 

87.  Since  a  may  represent  one  quantity  and  b  any  other 
quantity,  it  is  evident  that  any  trinomial  having  the  form 
#  a  +  2ab  -\-  b*  or  a*  —  %ab  -f-  £2  is  a  perfect  square.  • 

Rule.  —  Extract  the  square  roots  of  the  first  and  the  last 
term  of  the  trinomial,  and  connect  the  results  by  the  sign  of 
the  second  term. 

EXAMPLE.  —  Factor  _r2  4-  %xy  +y*. 

SOLUTION.  —  We  first  see  if  the  trinomial  has  the  form  stated  in  Art. 
86.  The  first  and  the  last  term  we  see  to  be  perfect  squares,  and  their 
roots  to  be  x  and^.  The  second  term  is  also  twice  the  product  of  the 
roots  x  and  y,  and,  since  it  has  the  plus  sign,  the  binomial  root  must  be 
x  +y.  Hence,  we  have  a  square  of  the  form  a?  +  %ab  -f-  b*,  and 

x"1  +  2xy  +y  =  (x  +y)  (x  +y)  =  (x  +y)\     Ans. 

EXAMPLE.—  Factor  16w4  +  9«6  —  24w2«3. 

SOLUTION.  —  The  first  term  of  the  expression  is  a  perfect  square,  but 
the  last  term  is  not.  Inspecting  the  second  term,  we  find  it  to  be  the 
square  of  3«3,  and  the  third  term  to  be  twice  the  product  of  3#3  and 
the  square  root,  4w2,  of  the  first  term.  Arranging  the  trinomial 
so  that  the  first  and  the  last  term  are  perfect  squares,  we  get 
Wm*  —  24m*n3  +  9«6  (a  square  of  the  form  fl2  —  2ab  -+-  £2),  and  we  have 


=  (4;«2-3«3)2.     Ans. 

EXAMPLE.  —  Factor  4x2  +  x^y"1 

SOLUTION.  —  Arranging  the  trinomial  so  that  the  first  and  the  last 
term  are  perfect  squares,  we  have  4,r2  +  2.r2_y  +  .i'2^2.  Now,  although 
the  first  and  the  last  term  are  perfect  squares  with  roots  Zx  and  xy, 
respectively,  the  second  term  is  only  equal  to  the  product  of  the  roots  ; 
hence,  the  trinomial  is  not  a  perfect  square,  and  can  only  be  factored 
by  Case  I.  Each  term  contains  .r2,  and  we  have  4.i  -2  +  x*y* 
=  .r24  +  +  2.  Ans. 


32  ELEMENTARY  ALGEBRA  AND  §  3 

88.  Should  two  of  tlie  terms  of  a  trinomial  be  perfect 
squares,  and  have  like  signs,  and  the  other  term  be  twice  the 
product  of  their  roots,  the  trinomial  is  a  perfect  square. 

Compare  this  statement  with  Art.  86.  Thus,  %ab  —  a*  —  b'\ 
if  divided  by— 1,  becomes  —  2at>  -f-  a2  +  b*  =  a*  —  2ab  +  b* ; 
hence,  2al>-a*-l>*  =  -  (a* - Zab  +  b*}  =  -(a-b)\ 

EXAMPLE. — Factor  ±pq  —  4/2  —  q'\ 

SOLUTION. — Dividing  first  by  —  1  we  have  —  4/ty+4/3  +  0rS  =  4/2 
—  ±pq  +  ?*  •—  (Zp  —  q}2.  Hence,  ±pq  —  4/2  —  q'2  —  —  (4/2  —  ±pq  +  q*) 
=  -(%p-qy.  Ans. 

EXAMPLE.— Factor  16r2.?2  +  16r4  +  4s4 . 

SOLUTION. — The  expression  contains  three  squares,  but,  by  careful 
inspection,  we  see  that  the  first  term  is  also  twice  the  product  of  the 
square  roots  of  the  other  two.  Thus, 

16rV  +  16r4  +  4^4  =  16r4  +  16r  V  +  4.y4  =  (4r2  +  2j2)2.     Ans. 


EXAMPLES  FOU  PRACTICE. 

8O.      Factor  the  following  trinomials: 

1.  -r*  -  16.r  +  64.  Ans.  (.r-8)2. 

2.  ;z6  -  26«3  +  169.  Ans.  («3  -  13)2. 

3.  25.r2  +  IQxyz  +  49y22-2.  Ans.  (5.r  +  lyz)*. 

4.  16^2  +  ^2-8^.  Ans.  (4c-W. 

5.  %mx  —  m*  —  ,r2.  Ans.  —  (m  —  .r)2. 

6.  «2^V6-2^V3  +  1.  Ans.  (ab*c*  -  I)8. 


CASE  III. 

9O.  To  factor  an  expression  which  is  the  difference 
between  two  perfect  squares: 

This  case  is  the  inverse  of  formula  3,  Art.  65,  and  may 
be  expressed  by  the  formula 

-b).  (6.) 


91.  Since  a  may  represent  one  quantity  and  b  any  other 
quantity,  it  is  evident  from  formula  6  that  any  expression 
which  is  the  difference  between  two  perfect  squares  may  be 
factored  by  the  following1 


g  3  TRIGONOMETRIC  FUNCTIONS.  33 

Ilule.  —  Extract  the  square  roots  of  the  first  and  the  last 
term.  Write  the  first  root  phis  the  second  for  one  factor,  and 
the  first  root  minus  the  second  for  the  other. 

EXAMPLE.  —  Factor  9.*^"  —  4. 

SOLUTION.  —  The  square  roots  of  the  first  and  the  last  term  are  3.t4y* 
and  2.  The  sum  of  these  roots  is  3.i'4j/3  +  2,  and  the  second  subtracted 
from  the  first  is  3.r4j/3  —  2.  Hence,  by  formula  6,  letting  a  =  8-r4j>3  and 
b  =  2, 

_  4  =  (3.*  V  +  2)  (3*y3  -  2).     Ans. 


EXAMPLE.—  Factor  (a  +  by  —  m'2n'\ 

SOLUTION.  —  The  square  roots  of  the  first  and  the  last  term  are  a  +  b 
and  mn.  The  sum  of  these  roots  is  a  +  b  +  mn,  and  the  second  sub- 
tracted from  the  first  is  a  +  b  —  mn.  Hence,  by  formula  6,  letting 
a  =  a  +  b  and  b  =  mn, 

(a  +  by  —  m^ri*  =  (a  +  b  +  mn)  (a  +  b  —  mn).     Ans. 


EXAMPLES  FOR  PRACTICE. 

9£.      Factor  the  following  expressions: 

1.  a'2  —  16.  Ans.  (a  +  4)  (a  —  4). 

2.  a'2  —  49<r8.  Ans.  (a+lc*)(a  —  lc*). 

3.  8Lr'>4-l.  .    Ans.  (9.r3/2  +  1)  (9.ry  -  1). 

4.  (ax  +  by)'*  —  1.  Ans.  (ax  +  by  +  1)  (ax  -\-by-  1). 

5.  25.r4/J  -  (bx  +  I)2.  Ans.  [5.r2j  +  (bx  +  1)]  [5*y  -  (bx  +  1)] 

=  (5*y  +  bx  +  1)  (5*y  -bx-  1). 

6.  l-169.rV.  Ans.    1 


93.  In  example   5,  the  expression   (^r-fl)2  should  be 
regarded  as  a  single  term  ;  in  fact,  any  number  of  terms  may 
be  regarded  as  a  single  term  by  enclosing  them  in  parenthe- 
sis and  operating  on  them  as  though  they  were  a  single 
letter. 

94.  When  solving  any  examples  requiring  the  applica- 
tion of  the  rule  in  Art.  91,  first  ascertain  if  the  numerical 
coefficients  of  the  two  terms  are  perfect  squares  ;  if  not,  there 
is  no  use  of  examining  further. 


34  ELEMENTARY  ALGEBRA  AND  §  3 

FRACTIONS. 


DEFINITIONS. 

95.  A  fraction,  in  algebra,  is  considered  as  an  expression 
indicating  division.     The  sign  -i-  is  seldom  used,  it  being 
more  convenient  to  write  the  dividend,  or  quantity  to  be 
divided,  above  a  horizontal  line,  with  the  divisor  below  it, 
in  the  form  of  a  fraction. 

Thus  the  fraction          .  means  that  a  -\-  b  is  to  be  divided 
c  —  d 

by  c  —  d,  and  is  the  same  as  (a-\-b)  -f-  (c  —  d).  It  is  read 
*  *  a  +  b  divided  by  c  —  */"  or  "  a  +  b  over  c  —  d."  All  fractions 
are  read  in  this  way  in  algebra,  except  simple  numerical 
fractions,  as  ^-,  -|f ,  etc. ,  which  are  read  as  in  arithmetic. 

96.  The  quantities  above  and  below  the  line  are  called 
the  numerator  and  the  denominator,  respectively,  as  in 
the  case  of  numerical  fractions.     They  are  known  as  the 
terms  of  a  fraction. 

97.  Since  dividing  any  quantity  by  1  does  not  change  its 
value,  we  may  write  any  quantity  as  a  fraction  by  making 
the  quantity  itself  the  numerator,  and  1  the  denominator. 

Thus,  7*y  may  be  written  — -^-,  and  not  be  altered  in  value. 

98.  A  reciprocal  of  a   quantity  is  1  divided  by  that 
quantity.     It  is  not  necessarily  written  as  a  fraction,  but 
when  so  written  has  the  quantity  for  the  denominator,  and 
1  for  the  numerator.     Thus,  the  reciprocal  of  5  is  £,  but  the 

reciprocal  of  £  is  ^  =  5;  the  reciprocal  of  -r'+j2  =     2        a; 

and  the  reciprocal  of  g  =  — !>••     Hence,  the  recipro- 

cal of  a  fraction  may  be  obtained  by  inverting  the  fraction. 

99.  The  three  signs  of  a  fraction  are :  the  sign  before 
the  dividing  line,  which  affects  the  entire  fraction ;  the  sign 
of  the  numerator;  and  the  sign  of  the  denominator.     When 
any  one  of  these  signs  is  omitted,  it  is  understood  to  be  plus. 


3  TRIGONOMETRIC  FUNCTIONS.  35 


Any  t^tvo  signs  of  a  fraction  may  be  changed  without  altering 
its  value,  btit  if  any  one,  or  all  three,  be  changed,  the  value 
of  the  fraction  will  be  changed  from  -\-to-  or  from  —  to  -j-. 

When  either  the  numerator  or  the  denominator  has  more 
than  one  term,  it  should  be  enclosed  in  a  parenthesis  when 
performing"  operations  affecting  it  as  a  whole.  The  paren- 
thesis may  be  removed  after  the  operations  are  completed. 

Take  the  fraction  -       —  ^  ;  placing  numerator  and  denom- 

inator in  parentheses,  we  have  —  -  --  -J-.     The  signs  of  the 

numerator  and  denominator  are  each  -f  and  that  of  the 
fraction  —  . 

Let  the  quotient  oia  —  b-^-c  —  d  =  q\  then  : 

' 


=-<+*>=-*• 


REDUCTION    OF   FRACTIONS. 

1OO.     To  reduce  a  fraction  is  to  change  its  form  without 

changing   its   value.      Thus,   — —    and    — —    have   different 

o  1U 

forms,  but  like  values,  since  10^-^5  and  20^-^10  are  each 
equal  to  2;r. 


36  ELEMENTARY  ALGEBRA  AND  §  3 

The  terms  of  a  fraction  may  botJi  be  multiplied,  or  may 
both  be  divided  by  the  same  quantity  without  changing  tJieir 
value. 

1O1.     To  reduce  a  fraction  to  its  simplest  form: 
Rule.  —  Resolve  each  term  into  its  factors,  and  cancel  fac- 
tors wJiidi  appear  in  both. 

1  O2.  In  per  forming  all  operations  on  fractions,  the  student 
must  learn  to  use  a  polynomial  factor  as  a  single  quantity,  like 
a  monomial  factor. 

This  is  illustrated  in  the  following  examples,  where  there 
are  polynomial  factors  in  both  numerator  and  denominator 
that  can  be  canceled. 

t2  +  2xy  +  y2 

EXAMPLE.  —  Reduce  -  -  0      ,         to  its  simplest  form. 
x  —y 

SOLUTION.  —  Factoring  both  numerator  and  denominator, 
y'z  _  (x  +y)  (x 


Canceling  the  common  factor  x  +y  from  both,  gives  as  the  result, 


3r5  —  6  rV 

EXAMPLE.  —  Reduce  77-  •..«     ,  to  its  simplest  form. 

22  8 


-  — 

SOLUTION.  —    •=—  =          ../  .  =  ^  —   ;  -  ~,  when  factored. 

§xy\x  —  2y) 


Canceling  the  common  factors,  we  have  as  the  result, 


__   *? 
~        ' 


2 


1O3.  Sometimes  the  whole  numerator  is  contained  in 
the  denominator,  or  the  denominator  in  the  numerator. 
The  numerator  or  denominator  will  then  reduce  to  the 
number  1. 

EXAMPLE.—  Reduce       +   ^  a  to  its  simplest  form. 

=      Ans' 


TRIGONOMETRIC  FUNCTIONS.  37 


EXAMPLE.  —  Reduce  -^  —  -  to  its  simplest  form. 


_  ..- 

SOLUTION.—    ^g-^i  =  -  —  ^,-»     i  -   =      i      =  •*+!•     Ans. 

(Art.  97.) 

1O4.     From  the  last  example  it  will  be  seen  that  division 

may  sometimes  be  performed  by  cancelation.     Thus,  —  — 

x  —  1 

means    (x*  —  1)  -j-  (-r3  —  1),    and    the  divisor  x*  —  1   canceled 
from  the  dividend  x*  —  1  gives  the  quotient  x*  -f-  1. 

A  factor  must  be  common  to  each  term  of  the  numerator 
and  to  each  term  of  the  denominator  in  order  to  be  canceled. 

Thus,  the  factor  x  cannot  be  canceled  from  -  -  —  because 

x  +  4;// 

it  is  not  common  to  both  terms  of  the  denominator. 


EXAMPLES  FOR  PRACTICE. 

1O5.      Reduce  the  following  to  their  simplest  form 

•.* 

Ans. 


a"2  —  b-'  'a  —  If' 

2.     '—z — ^-5.  Ans,  .r2  +}>"*. 

Qab^c 
Ans.  — -r — . 

Ans. 


- 
5.     -= -A — — -.  Ans. 


106.  When  fractions  are  to  be  added  or  subtracted,  it  is 
necessary  to  so  reduce  them  that  all  the  denominators  will 
be  alike.     This  is  called  reducing  them  to  a  common 
denominator. 

107.  To  reduce  fractions  to  a  common  denominator: 
Rule. — Resolve  each  denominator  into  its  factors. 

Take  each  factor  as  many  times  as  it  occurs  in  any  one 
denominator,  and  find  the  product  of  these  factors. 


38.  ELEMENTARY  ALGEBRA  AND  §  3 

Divide  this  product  by  each  of  the  denominators.  Multiply 
the  corresponding  numerators  by  these  quotients,  for  new 
numerators.  Write  each  new  numerator  with  the  common 
denominator  beneath  it. 

la         Zab  2b 

EXAMPLE.  —  Reduce  ,    >a  _    a,  and  -  -  —  to  a  common  denomi- 

nator. 

SOLUTION.  —  Factoring  the  denominators,  we  have  x+y  not  factora- 
ble. ,r2  —  y*  =  (.v  +y)  (x  —y),  and  (x  +  y)'*  —  (x  +y)  (x  +y).  Now  we 
have  two  separate  factors,  x  +y  and  x  —y,  of  which  x  +y  occurs  twice 
in  (x  +yY.  Hence,  our  common  denominator  is  (x  +y)  (x  +y)  (x  —y} 
—  (x  +yf  (x  —y}.  Dividing  this  product  by  x  +y  we  have  (x  +y) 
(x  —y)  =  x"*  —y"2  as  our  quotient.  Hence,  our  first  new  numerator  is 


1a(x*  -  y2)  and  the  new  fraction  is  ,  ~          .     Similarly, 

(x+yY(x-y)  *' 


becomes  -  \,  .  J  '  —  r,  and  ;  -  —  becomes 
' 


-  ,  .      —  r,          ;  -  —  -  -  -  ,      —  r. 

(x  +yY  (x  -y)'          (x  +yy  (x  +yY  (x  -y} 

The  student  should  note  that  this  denominator  can  be 
written  in  several  different  ways,  and  he  should  not  become 
confused  if  his  work  does  not  always  agree  with  the  answer. 
Besides  (x+y)  (x+f)  (x—y)  and  (x+y)'1  (x—y),  it  may 
be  written  (x*  —  /)  (x+y),  (x*  +  %xy  +  y*)  (x—y),  or 
x3  -f-  x*y  —  xy*  —y3.  These  five  expressions  have  exactly  the 
same  value.  The  student  should  .  prove  this  statement  by 
substituting  numbers  for  x  and  y. 


EXAMPLES  FOR  PRACTICE. 

1O8.      Reduce  the  following  to  a  common  denominator: 


2. 
8. 
4. 

R 

2x       ay                 42 

x*y     xy  z          .    lyz* 

-&<  is-  and  ir 

23                  4 

\2xyz      \Zxyz 
.         3.r2j/      2xyz 

Ans-  IT-  -W-- 
An-    3     Sa' 

12xyz* 

.    lax* 
and       „   „. 
a3x3 

—  2mn  +  «2 

a3x3'     ax3'    *     [    a'x 
m  +  n              m  —  n 

m  —  n               in  +  n 
2          3                    2x-l 

nnn 

m2  —  n1 

m'2  -  n2 

d 
'          L 


§  3  TRIGONOMETRIC  FUNCTIONS.  39 

ADDITION  AND  SUBTRACTION  OF  FRACTIONS. 

1O9.     To  add  or  subtract  fractions: 

Rule.  —  Reduce  the  fractions,  if  necessary,  to  a  common 
denominator.  Add  or  subtract  the  numerators,  and  write  the 
result  over  the  common  denominator. 

EXAMPLE.  —  Find  the  sum  of  —  =  —  and  —  -A  —  . 

5  4 

SOLUTION.  —    —  =  —  and  —  -.  —  ,  reduced  to  a  common  denominator, 
5  4 


4{2a  -b]       , 
become    v       —  -  and          —  -,  which  are  equal,  respectively,  to 

,*0  i*u  /& 

and  —  ^  —  .      Adding    the    numerators,    we    have    8a  —  4£  +  5a  +  §b 
=  13#  -|-  b.     The  result  written  over  the  common  denominator  gives  as 

The  work  is  written  as  follows: 
2a  —  b     a+b        Sa  —  ±b     5a  + 


the  sum,  — p^ — .     The  work  is  written  as  follows: 
20 


54  20  20 

Sa  -  ±b  -+-  5a  +  5b        ISa  +  b 


20 
EXAMPLE. — Subtract  —5-7 —  from  — - — . 

ou  4/tt 

SOLUTION. — Reducing  the  fractions  to  a  common  denominator, 


2<z 


-2        \2ab-Zb     \2ab-la 

Subtracting  the  second  numerator 


ab  §ab 

from    the    first,    and   writing  the    result  over  the    common   denom- 

\2ab-  3b        I2a&  —  4a          (IZab  —  3£)  —  (\2ab-  4a) 
inator,    we    have    — ^—7 — »— y —    =   * ~R~7, 

\2ab  —  %b  —  \2ab  +  4a 
=  —          —         ' ,  with  the  parentheses  removed.     Combining 

like  terms  in  the  numerator  gives  as  the  result          ,    .     Ans. 

HO.  If,  as  in  the  example  just  given,  the  numerator  of 
the  fraction  to  be  subtracted  has  more  than  one  term,  care 
must  be  taken  to  change  the  sign  of  every  term  before  com- 
bining. It  will  usually  be  convenient  to  enclose  the  whole 
numerator  in  a  parenthesis  before  combining.  The  paren- 
thesis may  then  be  removed  by  the  rules  of  Arts.  50 
and  51. 


40  ELEMENTARY  ALGEBRA  AND 


,.  0.       vr       **          *  x 

EXAMPLE. — Simplify  —  —  -\ 

J  x  —  1     x  +  \     x  —  \ 


SOLUTION. — Reducing  to  the  common  denominator  x1  —  1, 


Adding  or  subtracting  the  numerators  as  required,  we  have 
(.r4  +  ,r3)  -  (x3  -  .r2)  -  (x* 


which,  with  the  parentheses  removed,  equals 


Combining  like  terms  we  have  as  the  result 
'-— — -  —  .r2  +  l.     Ans. 


EXAMPLE.— Simplify    ^  +  - — 


SOLUTION. — If  the  denominator  of  the  second  fraction  were  written 
x  —  2  instead  of  2  —  x,  (x  —  2)2  would  be  the  common  denominator.  By 
Art.  99,  the  signs  of  the  denominator  and  the  sign  before  the  fraction 

may  be  changed,  giving = ; — ^.    (Art.  24.)    Hence, 


o "*~.r  «         ""fe^^)  &iTi"&       2  i    Y 

we   have  ; 55-=  +  s =  7 ^-r, -;,  which,  when  reduced  to  a 

(.f  —  i*)          &  — -  X          \X  —  tii)          X  —  & 

common  denominator,  is  equal  to 

1  x  —  2         1  —  (x  —  2)        1—  .r  +  2          3— .r 


EXAMPLES  FOR  PRACTICE. 

111.      Simplify  the  following  : 


xxx  A        47-t 

S+4+5-  AnS'6(r 

4.r-3     7o:  +  1     3^:  139.1-8 


8.     -    ____  --  Ans. 


A11<  -  _  (a  -  t>Y 

~  ~~~  ~~  ——'* 


removing  parentheses  and  combining. 
«2  a  a 


TRIGONOMETRIC  FUNCTIONS.  41 


—  - 

R  __  I  __  Arm 

4;;/2       "    12ma  "*      12;?    '  ' 

y          y         i 

' 


y(X  - 


xx  3.r  .r 

O.  5—        T.  .T\.llrt.  „ 


MUI/TIPLICATIOX  OF  FRACTIONS. 

Multiplication,  in  fractions,  is  the  process  of 
finding  a  fractional  part  of  a  fraction.  Thus,  fx£  means 
i  of  f.  One-half  of  f  inch,  for  example,  is  f  inch;  f  of 
f  inch  is  ^,  or  T62,  inch.  The  result  in  each  case  is  the  same 
as  that  which  would  be  obtained  by  finding1  the  product  of 
the  numerators  and  writing  it  over  the  product  of  the 
denominators. 

113.  Hence,  to  multiply  fractions: 

Rule. — Multiply  the  numerators  together  for  the  numera- 
tor of  the  product,  and  the  denominators  together  for  the 
denominator  of  the  product. 

114.  Any    number    of    fractions    may    be    multiplied 
together.     The  operation  may  be  very  much  shortened  by 
resolving  the  terms  of  the  fractions  into  their  factors,  and 
canceling.     The  product  should  be  reduced  to  its  simplest 
form. 

,  6rt2   Zab         ,  Zac 

EXAMPLE. — Find  the  product  of  -— ,  -^— ,  and  -75-. 

5       oc  b 

SOLUTION. — The  product  of  the  numerators  is  6«2  X  %ab  X  2ac  =  24a*frc, 
and  of  the  denominators,  5  X  &:  X  !>'*  —  150V.  Writing  Z±a*bc  over  150V. 

Z4a*bc     8a* 
we  have  for  the  product,  =  -^-,  when  reduced  to  its  lowest  terms. 

The  work  is  written  as  follows : 


42  ELEMENTARY  ALGEBRA  AND  §  3 

_  4  r  -I-  4- 


- 

EXAMPLE.—  Find  the  product  of    y  _     ,.  .*2  -  1,  and  - 

SOLUTION.  —  First  make  x*  —  \  a  fraction  by  writing  1  for  its  denom- 

x-i  _  i 
inator,  thus,  —  ^  —  ;   then,  factoring  both  terms  of  each  fraction,  we 


have 


*r  # 

EXAMPLE.  —  Find  the  product  of  —  =  --  ,  and  -  —  H  —  . 

«3       a  1  +  2ac 

I       4^ 

SOLUTION.  —  Performing  the  subtraction,  —  =  --  = 

'  a*       a 


EXAMPLES  FOR  PRACTICE. 

115.      Multiply  the  following: 

1.  •= — r~r>  by  -^ — -, — .  Ans.  • — . 
§abc           ¥>abc  c 

2.  •       •  by  2\xy.  Ans.  15^2_^2. 

Find  the  product  of: 

3^2y   Sy2^         ,  —  12jr2  .  15.r 

3-     i-rr2»  -flTr.  and    o-...2  •  Ans-  —  ^r^r- 


.r2  —  j/2      c  —  d  ,  .r3  +  y3  A         ^2  —  aj  +j/2 

4.     -T  —  ^.  -  r,  and  -  =  —  .  Ans.  -5  -  —,  —  ^t. 

)J  x  —y  c2  +  cd  +  d* 


4y      16  1 

5.     -*-  --  ,  and  -  --  j.  Ans. 

x       xy 


DIVISION  OF  FRACTIONS. 

116.  Division,  in  fractions,  is  the  reverse  of  multipli- 
cation, and  is  the  process  we  use  when,  given  one  of  two 
fractions  and  their  product,  we  are  required  to  find  the  other. 

For  example,  we  are  required  to  divide  -  by  -.     We  wish 


§  3  TRIGONOMETRIC  FUNCTIONS.  43 

to  find  such  a  fraction  that,    multiplied  by  -,  will  give  -. 

r      ,.        .     a    £      a     I        a       A1       x     x        7      . 
This  fraction  is  -,   for  ^X^  =  7.     Also  -=-^--^  =  ^  since 
A  &      Z         4:  5       7         0 

7     x        x 

-  x  -^  =  — .     If,  in  this  case,  we  had  inverted  the  divisor  and 

57         5 

x     7         7 

multiplied,  we  should  have  had  —  X  -  =  -. 

5     x        5 


117.     Hence,  to  divide  by  a  fraction:  , 

Rule.— 'Invert  the  divisor,  and  proceed  as  in  multiplication. 

EXAMPLE. — Divide  — -^-  by  ~ 

5.i-y   J  w.i-y •* 

SOLUTION. — The  divisor  inverted  = 


. 


.  _  _ 

Hence'  5P?^IOP?  =  5J>X"9^r  -  XX^>y      "8^ 

3  £' 

EXAMPLE.  —  Divide  ^2  -4-  2_t  +  1  by  "*  ' 

17      — 


SOLUTION.—  By  Art.  97,  (.r'  +  S.t+l)  - 

,T  —  T 

l)  _ 

-^-1.     Ans. 


EXAMPLES  FOR  PRACTICE. 

118.      Divide  the  following: 

I-       ^^  by  *f .  Ans.  5^L'. 

_      ab  —  bx  .      ac  —  ex  b 

2.                -  by  .  Ans.  -. 

a  +  z            a  +  z  c 

a     l-^  +  16^by  j_-4g^  Ans.  8«(1-2J). 


4.     ^cPcd  —  §abcd  by  — = ^ — 77,.  Ans.  a?  —  b*. 

7  a*  +  aA+6* 


44  ELEMENTARY  ALGEBRA  AND  §  3 

MIXED  QUANTITIES  AND  COMPLEX   FRACTIONS. 

119.     An  integral  expression  is  one  containing  neither 
fractions  nor  negative  exponents.     The  expression  c? 


1  3 

is  integral,  but  the  expressions  «2-f  ~—  -„  %a~*,     a      ,  are  not. 


o 
The  expression  2<?~2  is  only  another  way  of  writing  —  . 

The  use  of  negative  exponents  will  be  explained  in  subse- 
quent paragraphs. 

12O.     The  integral  part  of   an  expression  is  that  part 
which,  if  taken  by  itself,  would  be  an  integral  expression. 

121*     A  mixed  quantity  is  an  expression  containing 
both  integral  and  fractional  parts,  as  2#2  —       —  .     Consider- 

ing the  integral  part,  2<?2,  as  a  fraction  with  a  denominator  1 
(Art.  d7),  a  mixed  quantity  becomes  simply  the  indicated 
addition  or  subtraction  of  two  fractions;  thus, 


122.  A  fraction  may  be  reduced  to  either  an  entire  or 
mixed  quantity  by  dividing  the  numerator  by  the  denomi- 
nator, provided  the  division  be  possible.  It  frequently  hap- 
pens that  by  performing  the  indicated  division,  the  fraction 
will  be  reduced  to  a  simpler  form.  The  case  of  reducing  a 
fraction  to  an  entire  quantity  was  taken  up  in  Art.  1O3. 

f 
EXAMPLE.—  Simplify 


—  ^  -  . 

6X  -f-  o 

SOLUTION.  —  Performing  the  indicated  division, 

2.r  +  3  )  4.ra  +  12.r  -    1  (  2x  +  3  -  .  10  0.     Ans. 
4.r2+   6.t- 

6,1  -   1 

6.r+   9 

-10 

123.     Mixed  quantities  are  frequently  more  convenient 
to  handle  as  fractions. 


§  3  TRIGONOMETRIC  FUNCTIONS.  45 

To  reduce  a  mixed  quantity  to  a  fraction: 

Rule.  —  Write  the  integral  part   with   a   denominator  1, 
and  perform  the  indicated  addition  or  subtraction. 

EXAMPLE.  —  Reduce  x*  +  xy  +^2  --  •    —  to  a  fraction. 


SOLUTION.  —    x^  +  xy+y^  ---  =  -    —  •£  —  -  ---   —  ;  subtract- 

x  —y  \  .v  —  y 

ing  the  second  fraction  from  the  first  gives 


x  -y  x  -y 


EXAMPLES  FOR  PRACTICE. 

124.      Solve  the  following  : 

1.     Reduce  —      —  to  a  mixed  quantity.  Ans.  «2  H  --  . 


2.     Simplify  .  --.  Ans.  .r  +  8,  -  8  + 


3.     Reduce  x  +  3  — to  a  fraction.  Ans. 


a  +  b  a  —  b  2a       2a(d+l) 

4.     From3rt  +  —  ~-  subtracts  --  -=-.      Ans.  2«  +  -    =       ^ 


~  .  .  ,       . 

a  a  ad 

2n  2n  m  +  n 

5.     Divide  m  +  n  --  by  m  —  n  --    —  .  Ans.  —    —  . 

m  —  n    •  ?n  +  n  m  —  n 

SUGGESTION.  —  First  reduce  the  mixed  quantities  to  fractions. 

125.  A  complex  fraction  is  one  which  contains  frac- 

0  +  -   a_j} 
tions  in  one  or  both  of  its  terms.     Thus,  ^,  —  :  —  ,  and 

7 
a 

—  are  complex  fractions. 
~d 

126.  Complex  fractions  can  be  reduced  by  performing 
the    indicated  -division;    thus,    |^  =  |^-f=i|xf  =  |.     A 
much  simpler  way  is  to  multiply  both  terms  by  the  least 


46  ELEMENTARY  ALGEBRA  AND  §  3 

common   denominator   of   the   fractions   contained.      Thus, 

5  v  £ 


f  X8  " 

127.     Hence,  to  simplify  a  complex  fraction: 

Rule.  —  Multiply  both  terms  by  the  common  denominator  of 
the  fractional  parts. 


y     x 
EXAMPLE.  —  Simplify  -  -  . 

y     "x 

SOLUTION.  —  The  common  denominator  of  the  fractional  parts  is  xy. 
Multiplying  each  term  by  this,  we  have 


Ans- 


-~ 

The  multiplication  can  frequently  be  performed  mentally, 
without  writing  the  common  denominator,  at  the  same  time 
canceling  common  factors. 

EXAMPLE.  —  Simplify  -  . 


2a< 


SOLUTION. — This  is  the  case  of  a  complex  fraction  in   which   the 
denominator  is  itself  a  complex  fraction. 

First,  consider  the  part o~~iT- 


1  \-a 

Multiplying  both  terms  by  1  —  a,  we  have 

a(\  —  a)  _        a  — a2        _  a  —  a* 

(1  +  a)  (i  _  a)  +  2a'2  ~  1  -  a'2  +  2a2  ~  1  +  a2' 

The  fraction  thus  becomes r. 


Multiplying  both  terms   by  1  +  «2,  the  common   denominator,  we 
have 

1  +  «2  1  +  a*       . 

= s-^ • — ?  =  •= .     Ans. 

l  +  a2  +  a  —  a2        1  +a 


TRIGONOMETRIC  FUNCTIONS.  47 

EXAMPLES  FOR  PRACTICE. 

Simplify  the  following: 


Ans. 


S.     c-s.  Ans.  — 


3.     2 -  Ans. 


SUGGESTION. —    2|  means  2  +  |.      Hence,  for  the  numerator,  multi- 
ply-2  by  the  least  common  denominator  8,  and  add  7. 

1  4 

4.     = .  Ans.  „  .„..  .,. 


THEORY    OF    EXPONENTS. 

An  exponent  has  already  been  denned  in  Art. 
12.  In  addition  to  positive  integral  exponents,  as  there 
defined,  we  frequently  have  to  deal  with  negative  and  with 
fractional  exponents.  Thus,  in  the  quantities  a3,  a~\  and 
#?,  we  have  the  positive  integral  exponent  3,  the  negative 
exponent  —  3,  and  the  fractional  exponent  |.  The  rules  for 
positive  and  integral  exponents  apply  also  to  negative  and 
fractional  exponents. 

13O.  Since  letters  may  represent  numbers,  they  may  be 
used  for  exponents,  the  same  as  figures.  Thus,  aw  means 
that  a  is  to  be  taken  as  many  times  as  a  factor  as  there  are 
units  in  ;/,  or  aXaX a,  etc.  to  ;/  factors.  Such  exponents 
are  called  literal  exponents.  Fractional,  negative,  and 
literal  exponents  are  all  read  by  using  the  word  expo- 
nent. Thus,  a^,  a~\  am,  and  an  are  read,  "a,  exponent  |," 
"#,  exponent  minus  4,"  "a,  exponent  ;/  over  /;/,"  and 
**#,  exponent  ;/,"  respectively. 


48  ELEMENTARY  ALGEBRA  AND  §  3 

131.  A  fractional  exponent  is  an  expression  of  a  root 
or  of  a  power  and  a  root  combined.  The  numerator  of  the 
fraction  denotes  the  power,  and  the  denominator  denotes 
the  root.  Thus,  Va  may  be  written  a*.  Suppose  we  wish 
to  find  the  cube  root  of  a";  instead  of  writing  it  Va*  —  a\ 
we  may  write  it  a*  =  a*.  Hence, 

The  numerator  of  a  fractional  exponent  denotes  a  power, 
and  the  denominator,  a  root. 

m  m 

For  example,  a?  —  Va\  cl  =  ^V2;  xn  —  \xm;  1c*  = 
the  exponent  ™  applying  only  to  the  c. 


132.     The  meaning  of   negative    exponents   may   be 

illustrated  as  follows :     Let  it  be  required  to  divide  a"  by  a*. 

By  Art.  7O,  we  have,  —  =  «4~2  =  #2;  likewise,  a^-^-a1  =  — 
=  «2-2  =  a0-,   also,  a^a*  =  ^  =  a«-*  =  a~\  etc.  .     From 

this,  it  will  be  seen  that  ^  =  a~2;  but,  by  Art.  71,  a*  =  1, 
so  that  1  may  be  written  for  a°,  in  this  expression;    thus, 

—  =  —  =  a~\     Hence, 
a*        a* 

A  quantity  affected  with  a  negative  exponent  denotes  the 
reciprocal  of  the  same  quantity  affected  with  an  equal  positive 
exponent.  (Art.  98.) 


133.     Since  in   -^  =  #~2,   the  a~*  changes   to   a*   when 

placed   in   the   denominator,   we   may  state   the   following 
principle  : 

A  factor  may  be  changed  from  the  numerator  to  the  denom- 
inator, or  from  the  denominator  to  the  numerator,  if  the  sign 
of  its  exponent  be  changed. 


n~*          1          n          nb%    x~%          y 

For  example,  —j-  =  —5—3;  —  T-T  =  —  ;  •=  —  r  —  ~^T>   etc- 
ab        abn*    alrk         a      5y~l        5x% 

In  the  last,  the  positive  exponent  1  of  the  y  is  not  written. 


§  3  TRIGONOMETRIC  FUNCTIONS.  49 

EXAMPLE.  —  Express,  with  positive  exponents, 


SOLUTION.  —  Since  these  terms  may  be  taken  as  fractions,  with  1  for 
the  denominators,  we  have,  by  transferring  the  letters  with  negative 
exponents  to  the  denominators, 


Ans' 


134.  The  student  must  note  very  carefully  that  factors 
of  an  entire  term  only  can  be  changed  from  numerator  to 
denominator,  or  vice  versa,  and  that  when  thus  changed 
they  become  factors  of  the  whole  of  the  other  term.  Thus, 

in      _a  c~*  cannot  be  transferred  to  the  numerator  by 

merely  changing  the  sign  of  the  exponent.      The  exponent 
may,  however,  be  made  positive  by  multiplying  both  terms 


T  . 

'  thus'      -*  =  '    In         '  lf  we  trans- 


fer  the  c"*,  it  becomes    a/,   ,     ..,  c*  becoming  a  factor  of  the 
fty+a) 

entire  denominator. 

EXAMPLE.—  Clear   *y-*2rl  +  -jg-^-3^  ^'  of  negative   expo- 
nents. 

SOLUTION.  —  Treat  each  term  of  the  expression  separately.     x*y-*z~* 
xiy-i2-\ 
=  --  j  -  ;  changing  the  factors  with  negative  exponents  to  the  denom- 

inator, and  at  the  same  time  changing  the  signs  of  the  exponents,  we 
~l  ^S  n0t  E  ^actor  °^  the  w^°le  denominator,  so 


y  v~l  — 

we  must  multiply  both  terms  of  the  fraction  by  the  reciprocal  of  y~l  or  y  ; 


of  the  entire  numerator,  so  we  write  them  as  factors  of  the  entire 
denominator,  with  the  signs  of  the  exponents  changed  ;  thus, 


'nce' 


50  ELEMENTARY  ALGEBRA  AND  §3 

EXAMPLE.  —Solve  the  following: 


2n 

+  c" 


_ 

Write  the  answers  with  positive  exponents. 
SOLUTION.—        rt'x^"1  =  rt3+(~'>   =  rt:t~J  =  a\     Ans. 
=  »'+<-*>  =  «'"*  =  «*.     Ans. 


9 

Ans. 


n— 2n        —n  _r^         ^ 


Ans. 


135.     A  letter  may  be  raised  to  any  power    by  multi- 
plying its  exponent    by   the  index   of    the   power.       Thus, 


(a*)i  =  a,  («')-*  =  a~\  etc. 

EXAMPLE.  —  Find    the     values    of    the     following:    (a  ')~i;  (r^2)l; 
(x*)-**-^-*)-*. 

SOLUTION.  —  In    the     first,    multiplying     the    exponents,    we    have 
—  IX—  I  =\-    Hence,  (flr1)"*  =  «*,  or  4/«.    Ans.      In   like   manner, 

=  c*rf-5,  Ans.,  since  1  X  f  =  |,  and  —  2  X  |  =  —  5. 
In  the  next  one,  (.r«)~b  =  ^~«6  and  (j;-f')~6  =  x?b.     Dividing, 


136.  A  root  of  a  letter  affected  with  an  exponent  is 
extracted  by  dividing  the  exponent  by  the  index  of  the  root. 
Thus,  Vy"  =  y™  —  j/12*4  =  y\  From  this  principle,  the  fol- 
lowing rule  may  be  deduced. 

To  extract  any  root  of  a  monomial  : 

Rule.  —  Extract  the  required  root  of  the  numerical  coeffi- 
cient, and  divide  the  exponent  of  each  letter  by  the  index  of 
the  root.  Make  the  sign  of  every  even  root  of  a  positive  quan- 
tity ±,  and  the  sign  of  every  odd  root  of  any  quantity  the 
same  as  that  of  the  quantity. 


§  3  TRIGONOMETRIC  FUNCTIONS.  51 


EXAMPLE.—  Find  the  value  of      2 


SOLUTION.  —  The  4th  root  of  256  is  4.  The  exponent  of  a  in  the  root 
is  4  -f-  4  —  1  ;  of  b,  12  -;-  4  =  3  ;  and  of  c,  8  -4-  4  =  2.  As  this  is  an  even 
root  of  a  positive  quantity,  the  sign  should  be  ±.  Hence, 

.     Ans. 


EXAMPLE. — Find  the  value  of 


SOLUTION.—  $27m3x'J  =  3w.r3;  -$V£Va  =  rts£V.  The  quantity 
is  positive,  and,  as  this  is  an  odd  root,  its  sign  must  be  the  samo,  or 
positive. 


Hence,  =  .     Ans. 


EXAMPLES  FOR  PRACTICE. 

137.      Clear  the  following  of  negative  exponents  : 

1.     jtV"2.^.  Ans. 


2, 


3.    : 

Express  the  following  without  radical  signs: 

4.      itfTr*.  Ans.  (£-2)*  or  < 

Find  the  values  of^  the  following  : 

6.  n?  X  m  "*.  Ans. 

7.  2a$  X  «  *£.  Ans.  2« 

w 

8.  ^ 2  -T-  |/^-».  Ans. 

9.  2,r~2  -j-  (,r2)"^  Ans.  2, 

10.     (a/~«J     X  W4"-  Ans.  ^ 


EQUATIONS. 


DEFINITIONS. 

138.     As  defined  in  Art.  5,  an  equation  is  a  statement 
of  equality  between  two  expressions,  as  x  +  0  =  14. 

Every  equation  has  two  parts,   called  the  first  and  the 


52  ELEMENTARY  ALGEBRA  AND  §  3 

second  member.  The  first  member  is  the  part  on  the  left 
of  the  sign  of  equality,  and  the  second  member  the  part 
on  the  right  of  that  sign.  In  x-\-6  =  14,  x-\-§  is  the  first 
member,  and  14  is  the  second  member. 

139.  Equations  usually  consist  of  known  and  unknown 
quantities ;  that  is,  of  quantities  whose  values  are  given, 
and  of  quantities  whose  values  are  not  given,  but  are  to  be 
found.     Thus,  in  x  +  6  =  14,  6  and  14  are  known  quantities, 
and  x  is  unknown  ;  but  since  by  the  statement  of  the  equa- 
tion, x-\-§  must  equal  14,  x  must  have  such  a  value  that 
when  added  to  6  the  sum  will  be  14.     Hence,  the  value  of  x 
is  fixed  for  this  particular  case,  and  in  a  similar  manner  the 
value  of  a  single  unknown  quantity  in  any  equation  is  fixed 
by  the  relations  that  it  bears  to  the  known  quantities,  and 
this  value  can  usually  be  found. 

140.  To  solve  an  equation  is  to  find  the  value  of  the 
unknown  quantity.     This  is  done  by  a  series  of  transforma- 
tions by  which  the  first   member   becomes   the   unknown 
quantity,  and  the  second  member  becomes  a  known  quantity, 
which  is,  therefore,  the  value  of  the  unknown  quantity. 


TRANSFORMATIONS . 

141.  In  transforming  an  equation,  the  equality  of  its 
members  must  be  preserved ;  otherwise  the  existing  rela- 
tions between  the  known  and  unknown  quantities  will  be 
destroyed.     Transformations  are  based  upon  the  following 
principles  : 

142.  In  any  equation  : 

I.  The  same  quantity  may  be  added  to  both  members. 
For  example,  if  2  be  added  to  both  members  of  x*  =  16,  the 
members  of  the  resulting  equation,  x*  -\-2  =  18,  will  be  equal. 

II.  The  same  quantity  may   be    subtracted  from    both 
members.     Thus,  if  .r2  =  16,  then  x*  —  2  =  14. 


§  3  TRIGONOMETRIC  FUNCTIONS.  53 

III.     Both  members  may  be  multiplied  or  both  divided  by 
the  same  quantity.     Thus,  if  x*  =  16,   then  2x*  =  32  and 


IV.  Both  members  may  be  raised  to  the  same  power. 
Thus,  if  x*  =  16,  then  x*  =  256. 

V.  Like  roots  of  both  members  may  be  extracted.     Thus, 
if  x*  =  16,  then  x  =  4. 

A  little  thought  will  show  that  none  of  these  operations 
will  destroy  the  quality  of  the  members.  In  the  equation 
16  =  16,  for  example,  by  I,  16  +  2  =  16  -f  2  ;  by  II,  16-2 
=  16  —  2  ;  by  III,  16x2  =  16X2,  etc.  It  is  to  be  observed, 
however,  that  after  any  transformation,  the  members  do  not 
equal  their  original  values. 

143.  To  transpose  a  term  in  an  equation  is  to  change 
it  from  one  member  of  an  equation  to  the  other.     A  term 
may  be  transposed  to  the  other  member  of  an  equation,  if  its 
sign  be  changed.     Thus,  in  the  equation  2^  +  5  =  13,  let  it 
be  required  to  transpose  +  5  to  the  second  member;  chang- 
ing its  sign,  we  have  %x  =  13  —  5,  or  2;r  =  8.     For,  subtract 
5  from  both  members  and  we  have  2;r  +  5  —  5  =  13  —  5,  or 
%x  =  8.    Suppose  we  had  %x  —  5  —  13;  changing  the  sign  of 
—  5,  and  placing  it  in  the  second  member,  we  have  %x  =  18; 
for  2^—  5  +  5  =  13  +  5,  or  %x  =  18. 

144.  When  the  same  term  appears  in  both  members  of  an 
equation,  with  the  same  sign,  it  may  be  dropped  from  each. 
This  is  called   cancelation.     Thus,  in  x-{-a  =  16  -\-a  we 
may  cancel  a,   and  have  x  =  16,   for   subtracting  a  from 
both  members  x-\-a  —  a  =  16  -\-  a  —  a;  hence,  x  =  16.     In 
x  —  a  =  16  —  a,  x  —  a-\-a  =  \§  —  a-\-a,  and  x  =  16.     But, 
in  x  —  a  =  16  +  0,  the  #'s  will  not  cancel,   since  they  have 
different  signs. 

145.  Changing  Signs.  —  It  is  sometimes  desirable  to 
change  the  sign  of  a  term  in  one  of  the  members  of  an  equa- 
tion.    This  may  be  effected  by  multiplying  both  members 


54  ELEMENTARY  ALGEBRA  AND  §  3 

by  —  1.  This  gives  the  same  result  as  changing  the  signs  of 
all  the  terms  of  both  members;  thus,  jr-j-^  +  S  _  a  __  x  _  7 
may  be  changed  to  —  x  —  a  —  3  —  —a-\-x  +  1.  According 
to  Art.  142,  III,  this  transformation  does  not  destroy  the 
equality  of  the  members. 

14G.  Clearing  of  fractions  is  usually  necessary  before 
performing  any  operations. 

Rule. —  To  clear  an  equation  of  fractions,  multiply  each 
term  of  the  equation  by  the  common  denominator  of  all  the 
fractions. 

EXAMPLE. — Clear  of  fractions  x  -f  ^  +  -£  — f  =  10. 

246 

SOLUTION. — The  common  denominator  of  all  the  fractions  is  12; 
multiplying  each  term  by  12,  we  have 

12.r  +  6*  +  9*  -  4r  =  120. 

2*          1      3*  +  2 
EXAMPLE. — Clear  of  fractions =  o~~~2 — T- 

SOLUTION. — The  common  denominator  is  2(.r2  —  4);  multiplying 
through  by  this  we  get  4*(*  —  2)  =  (*2  —  4)  —  2(3*  +  2) ;  removing  the 
parentheses  this  becomes  4*2  —  8*  =  *2  —  4  —  6*  —  4. 

Where  a  fraction  is  preceded  by  a  minus  sign,  care  must  be 
taken  to  change  the  sign  of  every  term  of  the  numerator  when 
clearing  of  fractions.  (See  Art.  HO.) 


EXAMPLES  FOR  PRACTICE. 

Clear  the  following  equations  of  fractions: 


1.  ,r  +      +     =  16  _    .  Ans.  28*2  +  21^2  -f  20*  =  448*  -  56. 

2.  -A-"^  =  £  Ans.  3*-  6*  +  18  =  2a. 

4  A  O 

o         x  a-b 

O.         •  -  j  -  *    —    -  r  -  1. 

a—  b  a+b 

Ans.  ax  +  bx  —  a?x  +  Px  —  a?  —  2ab  +  b'*  —  a1  +  fr*. 

1  .    *          a  +  b  o         n      LI 

4.        —  ;  =  -  -.  --  —  .  Ans.  *  =  *a  —  a?  +  P. 

a  —  b        a  —  b         x 


§  3  TRIGONOMETRIC  FUNCTIONS.  55 

SIMPLE  EQUATIONS  WITH  ONE  UNKNOWN  QUANTITY. 

148.  Simple  equations,  when  reduced  to  their  simplest 
form,  contain  the  first  power  only  of  the  unknown  quantity. 
When  there  is  but  one  unknown  quantity,  it  is  usually  rep- 
resented by  x.     Numbers  and  the  first  letters  of  the  alphabet 
are  used  for  known  quantities. 

149.  To   solve   simple   equations   with    one    unknown 
quantity: 

Hule.^-C/ear  of  fractions.  Remove  all  signs  of  aggrega- 
tion. Transpose  all  terms  containing  tJie  unknown  quantity 
to  tJie  first  member;  all  others  to  tJie  second  member.  Com- 
bine the  first  member  into  one  term,  and  simplify  tJie  second 
member.  Divide  both  members  by  tJie  coefficient  of  tlie 
unknown  quantity. 

In  some  cases,  this  order  of  operations  may  be  changed  to 
advantage. 

150.  To  verify  the  result,  substitute  the  value  of  the 
unknown  quantity  in  the  original  equation,   which  should 
then  reduce  so  that  both  members  will  be  alike.     When  this 
occurs  the  equation  is  said  to  be  satisfied. 

EXAMPLE. — Solve  2.r  +  5  =  25  —  Sx. 

SOLUTION. — Transposing  the  unknown  quantities  to  the  first  member, 
wehave  2.r  +  3.r  +  6  =  25.. 

Transposing  the  known  quantities  to  the  second  member,  we  have 

2.r  +  3.t  =  25-5. 

Combining  like  terms,  5.r  =  20. 

Dividing  by  5,  x  =  4.     Ans. 

Now,  if  the  substituting  of  4  for  x  will  satisfy  the  original  equation, 
we  know  our  answer  is  correct.  Thus,  substituting  4  for  x>  we  have 

2X4  +  5  =  25-3X4. 
8  +  5  =  25-12. 
13  =  13. 

Hence  the  equation  is  satisfied,  and  our  result  is  correct. 


56  ELEMENTARY  ALGEBRA  AND  §  3 

EXAMPLE.—  Solve  16  —  x  —  \  7,r  —  \§x  —  (§x  —  8*  —  6,r)]  }  =  0. 
SOLUTION.  —  Removing  the  symbols  of  aggregation  (Art.  52), 

16  —  x  —  Ix  +  Sx  —  $x  +  Sx  —  §x  =  0. 
Transposing  16  to  the  second  member, 

—  x  —  lx  +  $x  —  %x  +  Sx  —  Qx  =  —  16. 
Combining  like  terms,  —  12x  =  —  16. 

Dividing  both  members  by     —  12, 

x  =  -if  =  li.     Ans. 


1       8-Gx     2(6.^  +  7) 
EXAMPLE.—  Solve  —  f—  +  -,  =  —  -  --  h        M        . 

A  4  0  o 

SOLUTION.  —  Simplifying  the  first  and  the  last  term  of  the  equation, 
it  becomes 

x  +  l  +  -  =  —  _  --  1  --  Z—  . 
454 

Clearing  of  fractions,  we  have 

20.r  +  20  +  5  =  32  -24.r  +  3(hr  +  35. 
Transposing,    2(Xr  +  24.*  —  30.*  =  32  +  35—20  —  5. 
Combining,  14;r  =  42. 

Dividing  by  14,  x  =  3. 

001 
EXAMPLE.—  Solve  -  --  -  --  h-  -  =.  =  0. 

1—  x     l  +  x     1  —  x* 

SOLUTION.  —  Clearing  of  fractions  by  multiplying  by  1  —  ;r2, 

=  0. 

=  0. 
Uniting  terms,  §x         =  —  2. 

x         —  —  .4.     Ans. 

NOTE.  —    0  multiplied  or  divided  by  any  number  rr  0. 

151.  literal  equations  are  those  in  which  the  known 
quantities  are  represented  partially  or  wholly  by  letters.  In 
solving  such  equations,  we  cannot  always  combine  the 
unknown  quantities  into  one  term. 

EXAMPLE.  —  Solve  %ax  —  %b  =  x  +  c  —  Sax. 

SOLUTION.  —  Transposing  the  terms  containing  the  unknown  quanti- 
ties to  the  first  member  and  the  remaining  terms  to  the  second  mem- 
ber, and  combining  like  terms, 


Factoring  Sax  —  x  to  bring  x  alone  in  the  first  member, 

(5a  -  l)x  =  Zb  +  c. 

The  coefficient  of  x  is  now  5^  —  1,  this  being  considered  as  one 
quantity. 

T\-   '  *'  3^  +  c       . 

Dividing  by  5^  —  1,       x  =        _    .     Ans. 


3 


TRIGONOMETRIC  FUNCTIONS. 


57 


PROOF.  —  Since  the  original  equation  is  equivalent  to  5a.i  —  x  =  %b  4-  c, 
it  will  be  sufficient  to  satisfy  this  equation.  Hence,  substituting  the 
value  of  x, 


Canceling  the  5a  —  1,  3£  -f  c  = 

3x  +  l        Sbx  — 
EXAMPLE.  —  Solve  -  ^-  =  -=-  - 


SOLUTION.  —  Clearing  of  fractions, 

(3*  +  1)  [b(x  +  !)-«]  =  (x  +  1)  (Zbx  -  2a  -K  c),  or 
bx(x  +  1)  -  3ax  +b(x  +  \)-a  =  3bx(x  +  1)  -  (2a  -  c}  (x  +  1). 
Canceling  3bx(x  +  1)  from  both  members, 

—  Sax  -\-bx  +  b  —  a  =  —  2ax  +  cx  — 
Transposing  and  uniting  terms, 

—  ax  +  bx  —  ex  =  —  a  —  b  +  c. 
Changing  signs  and  factoring, 

(a  —  b  +  c}x  =  a  +  b  —  c. 
a  +  b  —  c 


EXAMPLES  FOR  PRACTICE. 

5.      Solve  the  following: 

1.  16  — 3;t  =  13  —  Gx.  Ans.  x  =  —  1. 

2.  3(4.r  _  5)  +  6  =  1  +  2x.  Ans.  .r  =  1. 

3.  6(5  -2x)  =  Q-2(x-2).  Ans.  x  =  2. 

Ans.  x  =  60. 
Ans.  *  =  41. 
Ans.  *  =  —  7. 
Ans.  *  =  4£. 

> 

Ans.  ;r  = 


jr+1      ^r+4  . 

5.    — „ = —  =  lo  — 


6x     x  —  5x 

'     ^~~^ 7" 

O  OX  —  i 


5  _ 


3-2*. 


=  0. 


8.    2x  —  4a  = 


a*  —  3a  +  2 


Ans.  .r  = 


5a*  +  Wa 


SUGGESTION. — Transposing  the  second  term  to  the  second  member, 
ax  +  2x  _  a'2  +  4a  +  4:  _  (a  +  2)2 


5a 


58       ELEMENTARY  ALGEBRA  AND       §  3 

Multiplying  both  sides  by  5#, 

«,+„  =  **»+«)•. 

40 
Solving  for  x, 

- 
- 


afc*  +  ,r2)    ,   ax  .  .  c 

10.  —  —  •  -  -  =  ab  -\  --  .  Ans.  x  —  T. 

ex          c  b 


PROBLEMS  LEADING  TO   SIMPLE  EQUATIONS  WITH  ONE 
UNKNOWN  QUANTITY. 

153.  There  are  two  steps  in  the  solution  of  problems  by 
algebra : 

I. — The  relations  which  exist  between  the  known  and 
the  unknown  quantities,  that  is,  between  those  whose  values 
are  given  in  the  problem  and  those  whose  values  are  required, 
must  be  stated  by  one  or  more  equations.  This  is  called 
the  statement  of  the  problem. 

II. — The  resulting  equation  or  equations  must  be  solved, 
giving  the  values  of  the  required  quantities. 

It  will  thus  be  seen  that  by  the  algebraic  method,  the 
answer  to  a  problem  is  used  in  the  solution  and  operated 
upon  as  though  it  were  a  known  quantity,  which  is  one  great 
advantage  over  the  arithmetical  method. 

154.  The  ability  to   state  a  problem  by  means  of  an 
equation  depends  upon  the  ingenuity  of  the  operator  and 
his  ability  to  reason,   rather  than  upon  his  knowledge  of 
algebra.     No  definite  rule  can  be  given   for   making   the 
statement,  but  in  general,  where  there  is  only  one  unknown 
quantity  in  a  problem : 

Decide  what  quantity  it  is  whose  value  is  to  be  found. 
This  will  be  the  unknown  quantity,  or  the  answer.  Then 
represent  the  unknown  quantity  by  x  and  form  an  equation 
that  will  indicate  the  relations  between  the  known  and  the 
unknown  quantities  as  stated  in  the  problem. 

The  equation  will  also  indicate  the  operations  that  would 
be  performed  in  proving  the  statement  made  in  the  problem 


g  3  TRIGONOMETRIC  FUNCTIONS.  59 

were  the  answer  known.  Hence,  the  equation  may  often  be 
formed  by  noticing  what  operations  would  be  performed 
upon  the  answer  in  proving. 

155.  The  following  examples  are  illustrations  of  the 
statement  and  the  solution  of  algebraic  problems ;  they 
should  be  studied  carefully. 

EXAMPLE. — Find  such  a  number  that,  when  14  is  added  to  its  double, 
the  sum  shall  be  30. 

SOLUTION. — The  quantity  whose  value  is  required  is  the  number 
itself.  As  this  is  the  unknown  quantity,  let  x  =  the  number,  whence 
2x  must  be  double  the  number.  Now  the  problem  states  that  when 
14  is  added  to  double  the  number  the  sum  will  be  30.  In  other  words, 
when  14  is  added  to  2.r,  the  sum  will  be  30.  Hence,  the  statement  of 
the  problem  in  the  form  of  an  equation  is 

2x  +  14  =  30 ; 
whence,  solving,  x  =  8.     Ans. 

EXAMPLE. — Find  a  number  which,  when  multiplied  by  4,  will  exceed 
40  as  much  as  it  is  now  below  40. 

SOLUTION. — Let  x  —  the  required  number,  which,  when  multiplied 
by  4,  becomes  4.r.  According  to  the  conditions  of  the  problem,  the 
amount  by  which  4  times  the  required  number,  or  4.r,  exceeds  40  is 
equal  to  the  amount  that  the  number  itself,  or  x,  is  below  40. 

But  4x  —  40  is  the  amount  by  which  4x  exceeds  40,  and  40  —  x  is  the 
amount  by  which  x  is  below  40. 

Hence,  by  the  conditions,  we  have  the  statement, 
4.r-40  -  40 -.r. 

Transposing  and  uniting,       5,r  =  80, 
or  x  =  16.     Ans. 

EXAMPLE. — Two  loads  of  brick  together  weigh  4,000  Ib. ;  but  if 
500  Ib.  be  transferred  from  the  smaller  to  the  larger  load,  the  latter 
will  weigh  7  times  as  much  as  the  former.  How  much  does  each 
load  weigh  ? 

SOLUTION. — If  the  weights  of  the  two  loads  were  known  and  it  was 
desired  to  prove  the  correctness  of  the  example,  we  should  add  500  Ib. 
to  the  weight  of  the  larger  load  and  subtract  500  Ib.  from  the  weight 
of  the  smaller  load,  as  stated  in  the  example.  The  larger  load  should 
then  weigh  7  times  as  much  as  the  smaller.  To  obtain  the  equation 
the  same  thing  is  done  by  letting  x  =  the  weight  of  one  load,  whence 
4,000  —  .v  is  equal  to  the  weight  of  the  other  load. 

Let  x  —  the  weight  of  the  smaller  load. 


60  ELEMENTARY  ALGEBRA  AND  §3 

Then,  4,000  -  x  =  the  weight  of  the  larger  load. 

Also,  x  —  500  =  the  weight  of  the  smaller  load  after  transferring 
500  Ib. 

And  4,000  —  ^  +  500  =  the  weight  of  the  larger  load  after  transfer- 
ring 500  Ib. 

By  the  conditions,  the  larger  load  now  weighs  7  times  as  much  as 
the  smaller. 

Hence,         l(x  -   500)  =  4,000-.*  +  500. 

Solving,        7.*  -3,  500  =  4,500-jr, 
or  8x  =  8,000; 

whence,  x  —  1,000  Ib.  =  weight  of  smaller  load,  j     » 

and  4,000  -  x  =  3,000  Ib.  =  weight  of  larger  load,     i    ' 

PROOF.—  1,000  -  500  =  500  =  weight  of  the  smaller  load,  and 
3,000  +  500  =  3,500  =  weight  of  the  larger  load  after  the  500  pounds 
have  been  transferred  ;  3,500  -r-  500  =  7. 

Until  the  student  has  obtained  considerable  proficiency  in 
solving  problems  of  this  kind,  it  is  a  good  plan  to  prove  all 
problems. 

EXAMPLE.  —  The  circumference  of  the  fore  wheel  of  a  carriage  is  10 
feet,  and  of  the  rear  wheel,  12  feet.  What  distance  has  the  carriage 
traveled,  when  the  fore  wheel  has  made  8  more  turns  than  the  rear 
wheel  ? 

SOLUTION.  —  In  this  example  the  distance  traveled  is  not  known,  but 
is  required  to  be  found.  Suppose  that  the  distance  is  known,  and  that 
it  is  equal  to  x  feet,  and  that  we  wish  to  see  whether  the  statement  is 
true  that  the  fore  wheel  makes  8  more  revolutions  than  the  rear  wheel 
in  passing  over  x  feet.  The  number  of  revolutions  of  the  fore  wheel 

Jf  3C 

is  evidently  y^,  and  of  the  rear  wheel,  -^.     The  example  states  that 
the  difference  between  them  is  equal  to  8. 


Solving  for  x,  12*  -  10*  =  960, 

or  2x  =  960, 

and  x  =  480  feet.     Ans. 

480 
PROOF.  —        —  ^-  =  48  =  revolutions  of  fore  wheel. 

480 

—  —  -  =  40  =  revolutions  of  rear  wheel. 

48  —  40  =  8.     Compare  this  proof  with  (1). 

EXAMPLE.  —  A  water  cistern  connected  with  three  pipes  can  be  filled 
by  one  of  them  in  80  minutes,  by  another  in  200  minutes,  and  by  the 
third  in  300  minutes.  In  what  time  will  the  cistern  be  filled  when  all 
three  pipes  are  open  at  once  ? 


§  3  TRIGONOMETRIC  FUNCTIONS.  61 

SOLUTION.  —  Here  the  unknown  quantity  is  the  number  of  minutes 
required  to  fill  the  cistern  by  all  three  pipes  together.  Supposing  this 
to  be  x  minutes,  the  example  may  be  proved  by  noticing  that  the  sum 
of  the  quantities  of  water  flowing  through  each  pipe  separately  in  a 
given  length  of  time,  as  1  minute,  must  be  equal  to  the  quantity  flow- 
ing through  all  three  together  in  the  same  length  of  time.  According 
to  the  problem,  the  quantity  discharged  by  the  first  pipe  in  1  minute 
would  be  ^,  by  the  second  ^\^,  and  by  the  third  ^^  of  the  contents  of 
the  cistern.  In  like  manner  the  quantity  discharged  by  all  three  at 

once  in  1  minute  would  be  -.     Then,  if  the  example  is  stated  correctly, 
we  must  have 

-+  —  +  —  ~- 
80  "*"  200^300        x' 

Clearing  of  fractions,  ;r(30  +  12  +  8)  =  2,400, 
or  50.*  =  2,400; 

whence,  x  =  48  minutes.     Ans. 

EXAMPLE.  —  A  man  rows  a  boat  a  certain  distance  with  the  tide,  at 
the  rate  of  6|  miles  an  hour,  and  returns  at  the  rate  of  3i  miles  an 
hour,  against  a  tide  half  as  strong.  If  the  man  is  pulling  at  a  uniform 
rate,  what  is  the  velocity  of  the  stronger  tide  ? 

SOLUTION.  —  If  the  following  statement  is  not  clear,  the  student 
should  reason  it  out  for  himself  in  a  manner  similar  to  that  used  in  the 
last  three  examples. 

Let  x  =  number  of  miles  per  hour  that  the  stronger  tide  is  running  ; 

then  —  =  number  of  miles  per  hour  that  the  weaker  tide  is  running. 

& 

V 

Hence,  6|  —  x  and  3i  +  ^-  are  expressions  for  the  rate  at  which  the 

man  is  pulling.     But,  as  he  is  pulling  at  a  constant  rate  all  the  time, 
these  expressions  must  be  equal.     Hence, 


20  10     x 

or  -s-  -  x  —  -^  +  -. 

o  o       a 

Clearing  of  fractions,     40  —  6.r  =  20  +  3.v, 
or  —  9.r  =  —20; 

whence,  x  =  2|  miles  per  hour.     Ans. 


EXAMPLES  FOR  PRACTICE. 

156.      Solve  the  following  examples: 

1.     The  greater  of  two  numbers  is  four  times  the  lesser  number,  and 
their  sum  is  400 ;  what  are  the  numbers  ?  Ans.  80  and  320. 


62  ELEMENTARY  ALGEBRA  AND  §  3 

2.  A  farmer  has  108  animals,  consisting  of  horses,  sheep,  and  cows. 
He  has  four  times  as  many  cows  as  horses,  lacking  8,  and  five  times 
as  many  sheep  as  horses,  lacking  4 ;  how  many  has  he  of  each  kind  ? 

f  12  horses. 
Ans.   j  40  cows. 
[  56  sheep. 

3.  A  can  do  a  piece  of  work  in  8  days,  and  B  can  do  it  in  10  days ; 
in  what  time  can  they  do  it  working  together  ?  Ans.  4f  days. 

4.  Find  five  consecutive  numbers  whose  sum  is  150. 

Ans.  28,  29,  30,  31,  and  32. 

5.  A  boat  whose  rate  of  sailing  is  6  miles  per  hour  in  still  water 
moves  down  a  stream  which  flows  at  the  rate  of  3  miles  per  hour,  and 
returns,  making  the  round  trip  in  8  hours ;  how  far  did  it  go  down  the 
stream  ?  Ans.   18  miles. 


THE  TRIGONOMETRIC  FUNCTIONS. 


DEFINITIONS. 

157.  Plane   trigonometry  treats    of   the    solution    of 
plane  triangles. 

Every  triangle  has  six  parts — three  angles  and  three 
sides.  When  three  of  these  parts  are  given,  if  one  part  at 
least  is  a  side,  the  other  three  can  be  found.  This  process 
of  finding  the  unknown  parts  of  a  triangle  is  the  solution 
of  the  triangle. 

158.  The  complement  of  an  angle   is  the  difference 
between  90°  and  the  angle.     Thus,  the  complement  of  an 
angle  of  35°  is  an  angle  of  55°,  because  90°  -  35°  =  55°.      In 
a  right-angled  triangle,  the  right  angle  is  90° ;  since  the  sum 
of  the  three  angles  of  the  triangle  is  180°,  the  sum  of  the 
two  acute  angles  is  180°  —  90°  =  90°.     Therefore,  each  acute 
angle  of  a  right-angled  triangle  is  the  complement  of  the 
other  acute  angle. 

159.  The   supplement  of  an  angle  is  the  difference 
between  180°  and  the  angle.     Thus,  the  supplement  of  an 
angle  of  35°  is  an  angle  of  145°,  because  180°  —  35°  =  145°. 


g  3  TRIGONOMETRIC  FUNCTIONvS.  (>3 

160.  The   solution   of   a   triangle   is   accomplished    by 
means  of  the  trigonometric  functions.     These  functions 
are  the  ratios  of  the  sides  of  a  right-angled  triangle ;  the  niost 
important  of  these  functions  are  the  sine,  cosine,  tangent, 
and  cotangent.     These  are  abbreviated  to  sin,  cos,  tan,  and 
cot. 

161.  In   the   right-angled   triangle  ABC,   Fig.  2,  the 
sides  a,  b,  and  c  are  opposite,  respect- 
ively, to  the  angles  A,  B,  and  C.    The 

hypotenuse  is  c,  and  C  is  the  right 
angle.  The  short  side  b  is  adjacent  to 
angle  A  and  opposite  angle  B ;  the  short 
side  a  is  opposite  to  angle  A  and  adja- 
cent to  angle  B. 

Then  the  trigonometric  functions  are  defined  as  follows: 

162.  The    sine    of   an    angle    is   the   quotient,  of   the 
opposite  side  divided  by  the  hypotenuse.     Thus,  in  Fig.  2, 

sin  A  =  -c. 

163.  The  cosine  of  an  angle  is  the  quotient  of  the  adja- 
cent side  divided  by  the  hypotenuse.     Thus,  cos  A  =  -. 

164.  The   tangent    of   an    angle    is   the    quotient    of 
the   opposite   side   divided  by   the    adjacent   side.      Thus, 

tan  A  =  * 
b 

165.  The  cotangent  of  an  angle  is   the   quotient   of 
the   adjacent   side    divided   by   the   opposite   side.      Thus, 

cot  A  =  b-. 
a 


166.     From  the  definitions  of  the  functions,   we  have  : 

sin  B  =  -  ;  cos  B  =  -  ;  tan  B  —  -  ;  and  cot  B  =  7.     Com- 
c  c  a  b 

paring   the   functions   of   angle  B  with  those   of  angle  A, 


64  ELEMENTARY  ALGEBRA  AND  §  3 

sin  B  =  cos  A  (since  each  is  equal  to  -),  cos  B  =  sin  A, 

tan  B  —  cot  A,  and  cot  B  =  tan  A.  It  has  been  shown  (Art. 
158)  that  angle  A  is  the  complement  of  angle  B.  There- 
fore, the  sine  of  an  angle  is  equal  to  the  cosine  of  its  comple- 
ment, and  the  tangent  of  an  angle  is  equal  to  the  cotangent 
of  its  complement.  For  example,  sin  36°  =  cos  (90°  —  36°) 
=  cos  54°;  tan  28°  =  cot  62°;  etc. 


TRIGONOMETRIC    TABLES. 

167.  Every  angular  function  has  a  different  value  for 
each   of  the  angles   between   0°  and  90°.     The  numerical 
values  of  these  functions  are  called  natural  sines,  cosines, 
etc.,  and  are  given  in  the  tables  of  Natural  Sines,  Cosines, 
Tangents,  and  Cotangents.     In  many  tables,  both  natural 
and  logarithmic  functions  are  given.     The  student  should 
not  attempt  to  use  the  latter  until  he  thoroughly  under- 
stands logarithms.     The  table  of  natural  functions,  and  its 
use,  will  now  be  explained. 

168.  Given  an  Angle,  to  Find  Its  Functions. 

EXAMPLE. — Let  it  be  required  to  find  the  sine,  cosine,  and  tangent 
of  an  angle  of  37°  24'. 

SOLUTION. — Look  in  the  table  of  Natural  Sines  along  the  tops  of 
the  pages  and  find  37°.  The  left-hand  column  is  marked  ('),  meaning 
that  the  minutes  are  to  be  sought  in  that  column,  and  begin  with  0,  1, 
2,  3,  etc.,  up  to  60.  Glancing  down  this  column  until  24  is  found,  find 
opposite  this  24  in  the  column  marked  sine,  and  headed  37°,  the  number 
.60738;  then  .60738  =  sin  37°  24'.  In  exactly  the  same  manner,  find  in 
the  column  marked  cosine  and  headed  37°,  the  number  .79441,  which 
corresponds  to  cos  37°  24' ;  or  cos  37°  24'  =  .79441.  So,  also,  find  in  the 
column  marked  tangent  and  headed  37°,  and  opposite  24',  the  number 
.76456;  hence,  tan  37°  24'  =  .76456. 

169.  In  most  of  the  tables  published,  the  angles  run 
only  from  0°  to  45°  at  the  top  of  the  page;  to  find  an  angle 
greater  than  45°,  look  at  the  bottom  of  the  page  and  glance 
upivards,    using   the    extreme   right-hand    column    to 


§  3  TRIGONOMETRIC  FUNCTIONS.  65 

flml  minutes,  which  begin  with  0  at  the  bottom  and  run 
upwards,  1,  2,  3,  etc.  to  60. 

EXAMPLE. — Find  the  sine,  cosine,  and  tangent  of  77°  43'. 

SOLUTION. — Since  this  angle  is  greater  than  45°,  look  in  the  tables 
along  the  bottom  of  the  page,  until  the  column  marked  77°  is  found. 
Glancing  up  the  column  of  minutes  on  the  right,  until  43'  is  found,  find 
opposite  43'  in  the  column  marked  sine  (and  77°)  at  the  bottom,  the 
number  .97711 ;  this  is  the  sine  of  77°  43',  or  sin  77°  43'  =  .97711.  Sim- 
ilarly, in  the  column  marked  cosine,  find  opposite  43'  in  the  right-hand 
column,  the  number  .21275;  this  is  the  cosine  of  77°  43',  or  cos  77°  43' 
=  .21275.  So,  also,  find  that  4.59283  is  the  tangent  of  77°  43',  or 
tan  77°  43'  =  4.59283. 

170.  Let  it  be  required  to  find  the  sine  of  14°  22'  26". 
EXPLANATION. — The    sine    of   14°   22'   26"   lies  between 

sin  14°  22'  and  sin  14°  23'.  Sin  14°  22'  =  .24813;  sin  14°  23' 
=  .24841  ;  difference  =  .00028.  Neglect  for  the  moment 
the  fact  that  the  functions  are  decimal  fractions;  then  the 
difference  between  sin  14°  22'  and  sin  14°  23',  that  is,  the  dif- 
ference between  24841  and  24813,  is  28,  or,  corresponding  to  a 
difference  of  V  in  the  angle,  there  is  a  difference  of  28  in  the 
sine.  Now,  since  I'  =  60 ",  the  difference  between  sin  14°  22' 
and  sin  14°  22'  26",  that  is,  the  difference  corresponding  to 
a  difference  of  26"  in  the  angle,  must  be  f  f  X  28  =  12.1. 
Since  .1  is  less  than  .5,  omit  it,  and  we  have  12  as  the  differ- 
ence to  be  added  to  24813.  24813  +  12  =  24825;  taking 
account  now  of  the  fact  that  the  function  is  a  decimal  frac- 
tion, sin  14°  22'  26"  =  .24825. 

In  all  work  with  the  tables  it  will  always  be  found  most 
convenient  to  neglect  for  the  moment  the  decimal  point, 
consider  the  difference  a  whole  number,  and  afterwards 
locate  the  decimal  point  in  its  proper  position. 

171.  Reference  to  the  table  of  functions  shows  that,  as 
the  angles  increase  in  magnitude,   the  sines  and  tangents 
increase,  while  the  cosines  and  cotangents  decrease.     In  the 
above  example,  therefore,  had  it  been  required  to  find  the 
cosine  of  14°  22'  26",  the  correction  for  the  26"  would  have 
been  subtracted  from  the  cosine  of  14°  22 ',  instead  of  being 
added  to  it. 


60  ELEMENTARY  ALGEBRA  AND  §  3 

EXAMPLE. — Find  the  sine,  cosine,  and  cotangent  of  56°  43'  17". 

SOLUTION.— Sin  56°  43'  =  .83597.  Sin  56°  44'  =  .83613.  Since 
56°  43'  17"  is  greater  than  56°  43'  and  less  than  56°  44',  the  value  of  the 
sine  of  the  angle  lies  between  .83597  and  .83613 ;  neglecting  the  decimal 
character  of  the  functions,  the  difference  =  83613  —  83597  =  16;  multi- 
plying this  by  the  fraction  ^,  16  X  H  =  4-53-  Since  .53  exceeds  .5, 
we  take  5  as  the  difference  to  be  added  to  83597.  Adding, 
83597  +  5  =  83602;  hence,  sin  56°  43'  17"  =  .83602.  Ans. 

Cos  56°  43'  =  .54878;  cos  56°  44'  =  .54854;  expressed  as  a  whole 
number,  the  difference  =  54878  -  54854  =  24,  and  24  X  \l  =  7,  nearly. 
Now,  since  the  cosine  is  desired,  we  must  subtract  this  correction 
from  54878;  subtracting,  54878 -.7  =  54871.  Hence,  cos  56°  43'  17" 
=  .54871.  Ans. 

Cot  56°  43'  =  .65646;  cot  56°  44'  =  .65604;  difference  =  65646-65604 
=  42,  and  42  X  \\  —  12,  nearly.  Now,  since  the  cotangent  decreases 
as  the  angle  increases,  we  must  subtract  this  correction  from  65646 ; 
thus,  65646-12  =  65634.  Hence,  cot  56°  43'  17"  =  .65634.  Ans. 


172.  Given  the  Function,  to  Find  the  Correspond- 
ing Angle. — This  is  the  reverse  of  the  process  .for  finding 
the  function  of  the  angle.  If  the  angle  corresponding  to  the 
given  function  is  an  exact  number  of  degrees  and  minutes, 
the  function  will  be  found  in  the  table.  In  siich  a  case,  we 
have  simply  to  find  the  given  function,  and  take  the  degrees 
from  the  end  of  the  column,  and  the  minutes  at  the  end  of 
the  row.  If  the  name  of  the  function  is  at  the  top  of  the 
column,  the  number  of  degrees  will  be  found  there  also,  and 
the  minutes  at  the  left.  If  the  name  is  at  the  bottom,  the 
degrees  will  be  at  the  bottom,  and  the  minutes  at  the  right. 


CASE  I. 

173.      The  function  is  found  exactly  in  the  table. 
EXAMPLE. — Find  the  angle  whose  sine  is  .24982. 

SOLUTION. — In  the  table  of  Sines,  we  find  .24982  in  the  column  under 
14°.  Since  the  name  of  the  function  is  at  the  tqp  of  the  column,  we 
take  the  number  of  degrees  from  the  top,  and  the  minutes  from  the 
left;  thus,  14°  at  the  top,  and  28'  at  the  left.  Hence,  the  angle  whose 
sine  is  .24982  is  14°  28'.  Ans. 


§  3  TRIGONOMETRIC  FUNCTIONS.  67 

EXAMPLE. — Find  the  angle  whose  cotangent  is  .68557. 

SOLUTION. — In  the  table  of  Tangents,  we  find  .68557  under  34°.  The 
name  of  the  function  is,  however,  at  the  bottom  of  the  page,  and  we 
must  take  the  degrees  55°,  from  that  end  of  the  column.  The  minutes, 
34',  are.  found  in  the  right-hand  column.  Hence,  the  angle  whose 
cotangent  is  .68557  is  55°  34'.  Ans. 


CASE  II. 

174.     The  function  is  not  found  exactly  in  tJie  table. 

Let  it  be  required  to  find  the  angle  whose  sine  is  .42531. 

EXPLANATION. — Referring  to  the  table  of  Sines,  this  num- 
ber is  found  to  lie  between  .42525,  the  sine  of  25°  10',  and 
.42552,  the  sine  of  25°  11'.  Neglecting  decimals,  the  differ- 
ence between  these  two  sines  =  42552  —  42525  =  27 ;  the 
difference  between  42525,  corresponding  to  the  sine  of  25°  10', 
and  42531,  corresponding  to  the  sine  of  the  given  angle 
=  42531-42525  =  6.  Since  27  is  the  difference  for  1',  a 
difference  of  6  corresponds  to  -fa  of  1';  hence,  the  angle 
whose  sine  =  .42531  =  25°  10^'. 

Since  V  =  GO",  -/r  of  a  minute  =  ^8TX60  =  13.3".  There- 
fore, the  angle  whose  sine  is  .42531  =  25°  10'  13.3". 

The  given  function  should  always  be  compared  with  the 
function  of  the  angle  next  lo^ver,  and  the  correction  in  sec- 
onds should  be  added  to  that  angle.  In  the  case  of  the  sine 
and  tangent,  we  take  the  difference  between  the  given  func- 
tion and  the  next  smaller  function  appearing  in  the  tables; 
but  with  the  cosine  and  cotangent,  we  take  the  difference 
between  the  given  function  and  the  next  larger  function 
which  appears  in  the  tables. 

EXAMPLE. — Find  the  angle  whose  cosine  is  .27052. 

SOLUTION. — Looking  in  the  table  *of  Cosines,  the  given  function  is 
found  to  belong  to  an  angle  greater  than  45°,  and,  hence,  must  be  sought 
for  in  the  columns  marked  cosine  at  the  bottom  of  the  page.  It  is 
found  between  the  numbers  .27060  =  cos  74°  18'  and  .27032  =  cos  74°  19'. 
The  difference  between  the  two  is  .27060  —  .27032  =  .00028,  or  28,  neg- 
lecting decimals.  The  cosine  of  the  smaller  angle,  or  74°  18',  is  .27060, 
and  the  difference  between  this  and  the  given  cosine  is  .27060  — .27052 
—  .00008,  or  8,  neglecting  decimals. 

Hence,  /¥X60  =  17.14",  nearly,  and  the  angle  whose  cosine  is 
.27052  =  74°  18'  17.14",  or  cos  74°  18'  17.14"  =  .27052.  Ans. 


68  ELEMENTARY  ALGEBRA  AND  §  3 

EXAMPLE. — Find  the  angle  whose  tangent  is  2.15841. 

SOLUTION.—  2.15841  falls  between  2.15760  =  tan  65°  8',  and  2.15925 
=  tan  65°  9'.  The  difference  =  2.15925-2.15760  =  .00165,  or  165,  con- 
sidered as  a  whole  number.  2.15841  —  2.15760  =  81,  neglecting  the 
decimal.  Tyr  X  60  =  29.5",  nearly,  and  the  angle  whose  tangent  is 
2.15841  =  65°  8'  29.5",  or  tan  65°  8'  29.5"  =  2.15841. 


EXAMPLES  FOR  PRACTICE. 

175.      Solve  the  following  examples : 

1.  Find  the  (a)  sine,  (b)  cosine,  and  (c)  tangent  of  48°  17'. 

C  (a)    .74644. 
Ans.    •{  (b)     .66545. 
((c)     1.12172. 

2.  Find  the  (a)  sine,  (b}  cosine,  and  (c)  cotangent  of  13°  11'  6". 

•(•(a)    .22810. 
Ans.    J  (b)     .97364. 
[  (c)     4.26855. 

3.  Find  the  (a)  sine,  (b)  cosine,  and  (c)  tangent  of  72°  0'  1.8". 

f  (a)     .95106. 
Ans.    -j  (b)     .30901. 
[  (c)     3.07777. 

4.  (a)  Of  what  angle  is  .26489  the  sine,  (b)  of  what  is  it  the  cosine, 
and  (c)  of  what  is  it  the  cotangent  ?  (  (a)     15°  21'  37.2". 

Ans.    \  (b)     74°  38'  22.8". 
[  (c)     75°  9'  49". 

5.  (a)  Of  what  angle  is  .68800  the  sine,  (b}  of  what  the  cosine,  and 
(c)  of  what  the  tangent  ?  f  («)     43°  28'  20". 

Ans.    \  (b}     46°  31'  40". 
\(c)     34°  31' 40. 5". 


SOLUTION  OF  TRIANGLES. 


RIGHT-ANGLED  TRIANGLES. 

176.  When  any  side  and  one  of  the  acute  angles  of  a 
right-angled  triangle  (also  called  right  triangle)  are  given, 
the  remaining  sides  and  angles  may  be  found ;  also,  when 
any  two  sides  are  given,  the  remaining  parts  may  be  found. 


§  3  TRIGONOMETRIC  FUNCTIONS.  69 

177.  The  following  relations   between   the   sides   and 
angles  of   a   right   triangle   are  derived  directly  from  the 
definitions  of  the  trigonometric  functions: 

I.  Side  opposite  an  angle  =  hypotenuse  X  sine  of  angle. 

II.  Side  adjacent  =  hypotenuse  X  cosine. 

III.  Side  opposite  =  side  adjacent  x  tangent. 

IV.  Side  adjacent  =  side  opposite  X  cotangent. 

TT     TT  side  opposite 

V.   Hypotenuse  =  -  — . 

sine 

rj.  side  adjacent 

VI.   Hypotenuse  = 4 . 

cosine 

These  relations  are  sufficient  to  find  the  sides.  The  angles 
may  be  found  from  the  sides  by  the  relations  given  in  Arts. 
162  to  165.  To  show  the  application  of  these  relations  to 
the  solution  of  triangles,  a  number  of  examples  are  given. 

178.  Case  I. —  The  hypotenuse  and  an  acute  angle  being 
given,  to  find  the  remaining  parts : 

EXAMPLE. — In  Fig.  3,  the  hypotenuse  A  B  of  the  right-angled  tri- 
angle A  C  B  is  24  feet  and  the  angle  A  is  29°  31' ;  to  find  the  sides  A  C 
and  B  C  and  the  angle  B. 

NOTE. — When  working  examples  of  this 
kind,  construct  the  figure,  and  mark  the 
known  parts.  This  is  a  great  help  in  solving 
the  example.  Hence,  in  the  figure  draw  the 
angle  A  to  represent  an  angle  of  29°  31',  and 
complete  the  right-angled  triangle  A  C  B, 
right-angled  at  C,  as  shown.  Mark  the  angle  FlG-  3- 

A  and  the  hypotenuse,  as  is  done  in  the  figure. 

SOLUTION.— Angle  B  =  90°  -  29°  31'  =•  60°  29'.  A  Cis  the  side  adja- 
cent to  angle  A.  Hence,  from  relation  II,  Art.  177,  A  C,  or  side 
adjacent  =  hypotenuse  X  cosine  ='  24  X  cos  29°  31'.  In  the  table  of 
Natural  Cosines  we  find  the  cosine  of  29°  31'  to  be  .87021.  Therefore, 
A  C  =  24  X  .87021  =  20.89  feet,  nearly. 

To  find  the  side  B  C,  we  use  relation  I,  Art.  177.  B  C,  or  side 
opposite  =  hypotenuse  X  sine  =  24  X  sin  29°  31'.  The  sine  of  29°  31' 
is  .49268;  hence,  B  C  =  24  X  .49268  =  11.82  feet,  nearly. 

f  Angle  B  -  60°  29'. 
Ans.  J  Side  A  C  =  20.89  ft. 
[Side  BC  =  11.82ft. 


70  ELEMENTARY  ALGEBRA  AND  §  3 

179.  Case  II. — An  acute  angle  and  one  of  the  short  sides 
being  given,  to  determine  the  remaining  parts : 

EXAMPLE. — In  Fig.  3,  suppose  the  side  A  C  to  be  75  feet  and  the 
angle  A  to  be  32°  24'.  The  sides  B  C,  A  B,  and  the  angle  B  are 
required. 

SOLUTION.— Angle  B  =  90°  -  32°  24'  =  57°  36'.  To  find  B  C,  we  have 
relation  III,  Art.  177,  B  C,  side  opposite  =  side  adjacent  X  tangent 
=  75  X  tan  32°  24'.  Referring  to  the  table  of  Natural  Tangents,  the 
tangent  of  32°  24'  =  .63462.  B  C  =  75  X  .63462  =  47.6  ft,  nearly. 
To  find  the  hypotenuse  A  B,  we  use  relation  VI,  Art.  177,  hypotenuse 
_  side  adjacent  .  „  _  A  C  75  75  _  „ 

cosine  "  cos  A  ~  cos  32°  24'  ~  .84433  ~ 

f  Angle    B  =  57°  36'. 
Ans.  J  Side  B  C  =  47.6  ft. 
[  Side  A  B  =  88.81  ft. 

180.  Case  III. —  Two  sides  being  given,  to  find  the  third 
side  and  the  acute  angles : 

EXAMPLE. — In  the  right-angled  triangle  A  B  C,  Fig.  4,  right-angled 
B    at   C,  A  C  =  18  and  B  C  -  15;  to  find  A  B 
and  the  angles  A  and  B. 

SOLUTION. — According  to  the  definition  of 
the  tangent,  Art.  164, 

.        side  opposite        15 
tangent  A  =  -.  -  =  ~  =  .83333. 

side  adjacent        18 

Looking  in  the  table  of  Tangents,  the  angle 
whose  tangent  is  nearest  to  .83333  is  39°  48'. 
IS  ""     Hence,   angle  A  —  39°  48',  nearly.     Angle  B 

FIG.  4.  =  90°  _  39°  48'  =  50°  12'. 

To  find  the  hypotenuse,  we  use  relation  V,  Art.  177, 

side  opposite          B  C 

A  B.  hypotenuse  =  f£ =  -= -r. 

s*me  sm  A 

The  sine  of  39°  48'  is  .64011.     Hence,  A  B  =  — j^r  =  23.43. 


f  Angle  A  =  39°  48'. 
Ans.   J  Angle  B  -  50°  12'. 
[  Side  A  B  =  23.43. 

EXAMPLE. — In  the  right-angled  triangle  ABC,  Fig.  5,  right-angled 
at  C,  A  C  =  .024967  mile  and  A  B  —  .04792  mile ;  to  find  the  other  parts. 


TRIGONOMETRIC  FUNCTIONS. 


71 


SOLUTION. — According   to   the    definition  of 
the  cosine,  Art.  163, 

.        side  adjacent      '.024967 
COS  A  =      hypotenuse      =  1>4W  =  -53101- 
Referring    to    the    table,    the   angle   whose 
cosine  is  .52101  is  58°  36'.     Therefore,  angle  A 
=  58°  36'.     Angle  B  =•  90°  -  58°  36'  =  31°  24'. 
To  find  side  B  C,  relation  III,  Art.  177,  is 
used. 

Side  opposite  A  —  side  adjacent  X  tan  A, 

or  B  C  =  A  C X  tan  58°  36'  =  .024967  X  1.63826 

=  .0409  mile.  j'  Angle  A  =  58°  36'. 

Ans.  -j  Angle  B  =  31°  24'. 

\  B  C  =  .0409  mile. 


.024967 

FIG.  5. 


EXAMPLE.  —  In  the  right-angled  triangle  ABC,  Fig.  6,  right-angled 
at  C,  A  B  =.  308  feet  and  B  C  =  234  feet  ;  to  find  the  other  parts. 


SoLUT,ON.-Sin  A  = 


hypotenuse 


=  !*  =  .75974. 
308 


(Art.  163.) 


The  angle  whose  sine  is  .75974  =  49°  26|',  nearly,  =  angle  A. 
Referring  to  the  table  of  Sines,  the  sine  of  49°  26'  is  .75965,  and  the  sine 
of  49°  27'  is  .75984.  Since  the  value  obtained, 
.75974,  lies  nearly  half  way  between  these  val- 
ues, the  angle  lies  midway  between  the  above 
angles,  and  is  49°  261'.  Angle  B  =  90°  -  49°  261' 
=  40°  33|'. 

To  find  A  C,  use  relation  IV,  Art.  177. 
Side  adjacent  A  =  side  opposite  X  cot  A,  or 
A  C  =  234  X  .85586  =  200.27  feet. 

f  Angle  A  —  49°  261'. 
Ans.   J  Angle  B  =  40°  331'. 
[AC  =  200.27  ft. 


FIG.  6. 


EXAMPLES  FOR  PRACTICE. 
1 8 1 .      Solve  the  following  examples : 

1.     In  a  right  triangle  ABC,  right-angled  at   C,  the  hypotenuse 
A  B  =  40  inches  and  angle  A  =  28°  14'  14".     Solve  the  triangle. 

[Angle  B  =  61°  45' 46". 
Ans.   \  A  C  =  35.24  in. 
£C=  18.92  in. 


72  ELEMENTARY  ALGEBRA  AND  §  3 

2.  In  a  right  triangle  ABC,  right-angled  at  C,  the  side  B  C  =  10 
feet  4  inches.    If  angle  A  =  26°  59'  6",  what  do  the  other  parts  equal  ? 

(Angle  B  =  63°  0'  54". 
A  B  =  22  ft.  9£  in. ,  nearly. 
A  C  =  20  ft.  3|  in.,  nearly. 

3.  In  a  right  triangle  A  B  C,  right-angled  at   C,   the  hypotenuse 
A  B  =  60  feet  and  the  side  A  C  =  22  feet.     Solve  the  triangle. 

[  Angle  A  =  68°  29' 22. 2". 
Ans.  \  Angle  B  =  21°  30'  37.8". 
[£  C  =  55.82ft. 

4.  In  a  right  triangle  ABC,  right-angled  at  C,  side  A  C  =  .364 
foot  and  side  B  C  =  .216  foot.     Solve  the  triangle. 

f  Angled  =  30°  41' 7.5". 
Ans.  \  Angle  j9  =  59°  18'  52.5". 


i 


A  B  =  .423  ft. 


OBLIQUE-ANGLED  PLANE  TRIANGLES. 

182.  We  will  give   here   the  method   of   solving  any 
oblique-angled  plane   triangle,   when  (1)  two  sides  and  an 
angle  opposite  one  of  them  are  given,  and  (2)  when  two  angles 
and  a  side  opposite  one  of  them  are  given.      Let  the  student 
bear  in  mind,  however,  that  he  may  use  this  method  of  solu- 
tion only  when  he  knows  the  general  form  of  the  triangle  of 
which  he  desires  the  values  of  some  of  the  parts.     This  is 
necessary  because  in  some  cases  two  solutions  are  possible, 
resulting  in  the  determinations  of  triangles  of  quite  different 
forms.     We  do  not  think  it  necessary  to  go  so  deeply  into 
the  explanation  of  this  point  that  the  student  can  detect  the 
cases  where  two  solutions  are  possible. 

183.  The  solution  of  the  triangle  depends  upon  the  fol- 
lowing principle :  In  any  triangle,  the  sides  are  proportional 
to  the  sines  of  the  opposite  angles.     Thus,  referring  to  Fig.  7, 
the  following  proportions  are  true : 

a  :  b  —  sin  A  :  sin  B. 
a  \  c  —  sin  A  :  sin  C. 
b  :  c  =  sin  B  :  sin  C. 


§  3  TRIGONOMETRIC  FUNCTIONS.  73 

The  method  of  solving1  the  triangle  is  shown  in  the  follow- 
ing examples: 

184.  Case  I. —  Two  sides  and  an  angle  opposite  one  of 
them  are  given  : 

EXAMPLE. — In  the  triangle  ABC,    Fig.  7,  having  given  the   side 
a  =  1,686     feet,    the    side   b 
=  960  feet,  and  the  angle  A 
=  33°  35' ;  to  find  the  angle  B. 

SOLUTION. — We  have 

a  :  b  =  sin  A   :  sin  B. 
Substituting    the    known 
values,  we  have 

1,686  :  960  =  sin  33°  35'  :  sin  B. 

From  the  table  of  Natural  Sines  we  get  .55315  as  the  sine  of  33°  35'. 
Substituting  this  value  in  the  proportion  and  solving,  we  find  that 
.31496  is  the  sine  of  the  angle  B,  which,  by  consulting  the  table  of 
Natural  Sines,  we  find  to  correspond  nearly  with  the  angle  18°  22'. 

Ans. 

185.  Case  II. —  Two  angles  and  a  side  opposite  one  of 
them  are  given  : 

EXAMPLE. — In  the  triangle  ABC,  given  the  angle  A  =  33°  35',  the 
angle  B  =  18°  22',  and  the  side  a  —  1,686  feet;  to  find  the  side  b. 

SOLUTION. — We  have  sin  A  :  sin  B  =  a  :  b.  Substituting  known 
values,  we  get 

.55315  :  .31509  =  1,686  :  b. 

Solving  the  proportion,  we  find  b  =  960  feet.     Ans. 


TABLES 


OF 

NATURAL  SINES,   COSINES, 
|  TANGENTS, 

AND   COTANGENTS 

GIVING  THE  VALUES  OF  THE  FUNCTIONS  FOR 
ALL  DEGREES  AND  MINUTES  FROM 
.       O°    TO    QO° 


NATURAL  SINES  AND  COSINES. 


77 


j 

C 

o 

I 

0 

2 

0 

3 

o 

4 

0 

Sine 

Cosine 

Sine 

Cosine 

Sine 

Cosine 

Sine 

Cosine 

Sine 

Cosine 

0 

.00000 

•01745 

.99985 

.03490 

•99939 

.05234 

•99863 

.06976 

•99756 

60 

1 

.00029 

.01774 

.99984 

•03519 

.99938 

.05263 

.9986! 

.07005 

•99754 

59 

2 

.00058 

.01803 

•99984 

•03548 

•99937 

.05292 

.99?-6o 

.07034 

•99752 

58 

3 

.00087 

.01832 

.99983 

•°3577 

.99936 

.05321 

.99858 

.07063 

•9975° 

57 

4 

.00116 

.01862 

.99983 

.03606 

•99935 

•0535° 

.99857 

.07092 

.99748 

56 

5 

.00145 

.01891 

.99982 

•03635 

•99934 

•05379 

•99855 

.07121 

.99746 

55 

6 

.00175 

.01920 

.99982 

.03664 

•99933 

.05408 

.99854 

.07150 

•99744 

54 

7 

.  00204 

.01949 

.99981 

•03693 

•99932 

•05437 

•99852 

.07179 

•99742 

53 

8 

.00233 

.01978 

.99980 

•03723 

•99931 

.05466 

.99851 

.07208 

.99740 

52 

9 

.00262 

.02007 

.99980 

•03752 

.99930 

•05495 

.99849 

.07237 

•99738 

51 

10 

.00291 

.02036 

•99979 

•03781 

.99929 

•05524 

.99847 

.07266 

•99736 

50 

ii 

.00320 

•99999 

.02065 

•99979 

.03810 

•99927 

•05553 

.99846 

.07295 

•99734 

49 

12 

.00349 

.99999 

.02094 

.99978 

•03839 

.99926 

•05582 

.99844 

•07324 

•99731 

48 

13 

.00378 

•99999 

.02123 

•99977 

.03868 

.99925 

.05611 

.99842 

•07353 

.99729 

47 

14 

.00407 

•99999 

.02152. 

•99977 

.03897 

•99924 

.05640 

.99841 

.07382 

•99727 

46 

15 

.00436 

•99999 

.02181 

.99976 

.03926 

•99923 

.05669 

•99839 

.07411 

•99725 

45 

16 

.00465 

•99999 

.02211 

.99976 

•03955 

.99922 

.05698 

.99838 

.07440 

•99723 

44 

17 

.00495 

•99999 

.02240 

•99975 

.03984 

.  9992  i 

.05727 

.99836 

.07469 

.99721 

43 

18 

.00524 

•99999 

.02269 

•99974 

.04013 

.99919 

•05756 

.99834 

.07498 

.99719 

42 

*9 

•oo553 

.99998 

.02298 

•99974 

.04042 

.99918 

•05785 

•99833 

.07527 

.99716 

41 

20 

.00582 

•99998 

.02327 

•99973 

.04071 

.99917 

.05814 

.99831 

•07556 

.99714 

40 

21 

.00611 

.99998 

.02356 

•99972 

.04100 

.99916 

.05844 

.99829 

•07585 

.99712 

39 

22 

.00640 

.99998 

.02385 

•99972 

.04129 

•99915 

•05873 

.99827 

.07614 

.99710 

38 

23 

.00660 

.99998 

.02414 

.99971 

.04159 

•999!3 

.05902 

.99826 

.07643 

.99708 

37 

24 

.00698 

•9999s 

.02443 

.99970 

.04188 

.99912 

•05931 

.99824 

.07672 

•99705 

36 

25 

.00727 

•99997 

.02472 

.99969 

.04217 

.99911 

.05960 

.99822 

.07701 

•99703 

35 

26 

.00756 

•99997 

.02501 

.99969 

.04246 

.99910 

.05989 

.9982? 

.0773° 

.99701 

34 

27 

.00785 

•99997 

.02530 

.99968 

.04275 

.99909 

.06018 

.99819 

•07759 

.99699 

33 

28 

.00814 

.99997" 

.02560 

.99967 

.04304 

.99907 

.06047 

.99817 

.07788 

.99696 

32 

29 

.00844 

.99996 

.02589 

.99966 

•°4333 

.99906 

.06076 

.99815 

.07817 

•99694 

3i 

3° 

fc  .00873 

•99996 

.02618 

.99966 

.04362 

•99905 

.06105 

.99813 

.07846 

.99692 

3° 

31 

.  00902 

.99996 

.02647 

.99965 

.04391 

.99904 

•06134 

.99812 

•07875 

.99689 

29 

32 

.00931 

.99996 

.02676 

.99964 

.04420 

.99902 

.06163 

.99810 

.07904 

.99687 

28 

33 

.00960 

•99995 

.02705 

.99963 

.04449 

.99901 

.06192 

.99808 

•07933 

.99685 

27 

34 

.00989 

•99995 

•02734 

.99963 

.04478 

.06221 

.  99806 

.07962 

.99683 

26 

9 

.01018 
.01047 

•99995 
•99995 

.02763 
.02792 

.99962 
.99961 

.04507 
•04536 

•99897 

.06250 
.06279 

.99804 
.99803 

.07991 
.08020 

.99680 
.99678 

25 
24 

37 

.01076 

•99994 

.02821 

.99960 

•04565 

.99896 

.06308 

.99801 

.08049 

.99676 

23 

38 

.01105 

•99994 

.02850 

•99959 

.04594 

.99894 

•06337 

•99799 

.08078 

•99673 

22 

39 

.01134 

•99994 

.02879 

•99959 

.04623 

.99893 

.06366 

•99797 

.08107 

.9967: 

21 

40 

.01164 

•99993 

.02908 

.99958 

•04653 

.99892 

.06395 

•99795 

.08136 

.99668 

20 

4i 

.01193 

•99993 

.02938 

•99957 

.04682 

.99890 

.06424 

•99793 

.08165 

.99666 

19 

42 

.01222 

•99993 

.02967 

.99956 

.04711 

.99889 

•06453 

.99792 

.08194 

.99664 

18 

43 

.01251 

•99992 

.02996 

•99955 

.04740 

.99888 

.06482 

.99790 

.08223 

.99661 

17 

44 

.OI28o 

.99992 

.03025 

•99954 

.04769 

.99886 

.06511 

.99788 

.08252 

•99659 

!6 

45 

.01309 

.99991 

.03054 

•99953 

.04798 

.99885 

.06540 

.99786 

.08281 

•99657 

15 

46 

.01338 

.99991 

'.03083 

.99952 

.04827 

•99883 

.06569 

.99784 

.08310 

•99654 

14 

47 

.01367 

.99991 

.03112 

•99952 

.04856 

.99882 

.06598 

.99782 

•08339 

•99652 

13 

48 

.01396 

.99990 

.03141 

•99951 

.04885 

.99881 

.06627 

.99780 

.08368 

•99649 

12 

49 

.01425 

.99990 

.03170 

.99950 

.04914 

.99879 

.06656 

•99778 

.08397 

•99647 

II 

50 

.01454 

•99989 

.03199 

•99949 

.04943 

.99878 

.06685 

.99776 

.08426 

•'99644 

10 

Si 

.01483 

.99989 

.03228 

.99948 

.04972 

.99876 

.06714 

•99774 

•08455 

.99642 

9 

52 

•OI5I3 

.99989 

•03257 

•99947 

.05001 

•99875 

.06743 

.99772 

.08484 

•99639 

8 

53 

.01542 

.99988 

.03286 

.99946 

.05030 

•99873 

.06773 

.99770 

•08513 

•99637 

7 

54 

.01571 

.99988 

.03316 

•99945 

•05059 

.99872 

.06802 

.99768 

.08542 

•99635 

6 

55 

.01600 

.99987 

•°3345 

•99944 

.05088 

.99870 

.06831 

.99766 

.08571 

.99632 

5 

56 

.01629 

.99987 

•°3374 

•99943 

.05117 

.99869 

.06860 

•99764 

.08600 

.99630 

4 

57 

.01658 

.99986 

•03403 

.99942 

.05146 

.99867 

.06889 

.99762 

.08629 

.99627 

3 

58 

.01687 

.90986 

•03432 

.99941 

•05i75 

.99866 

.06918 

.99760 

.08658 

•99625 

2 

59 

.01716 

.99985 

.03461 

.99940 

.05205 

.99864 

.06947 

•99758 

.08687 

.99622 

I 

60 

•01745 

•99985 

.03490 

•99939 

.05234 

.99863 

.06976 

•99756 

.08716 

.99619 

O 

Cosine 

Sine 

Cosine 

Sine 

Cosine 

Sine 

Cosine 

Sine 

Cosine 

Sine 

/ 

r 

8 

?° 

8 

s° 

8 

7° 

8 

5° 

«. 

.0 

78 


NATURAL  SINES  AND  COSINES. 


5° 

6° 

7° 

8° 

9° 

f 

Sine 

Cosine 

Sine 

Cosine 

Sine 

Cosine 

Sine 

Cosine 

Sine 

Cosine 

0 

.08716 

.99619 

•  10453 

.99452 

.12187 

•99255 

.13917 

.99027 

•15643 

.98769 

~6o 

I 

.08745 

.99617 

.  10482 

•99449 

.  12216 

•99251 

•13946 

.99023 

.15672 

.98764 

59 

2 

•08774 

.99614 

.10511 

•99446 

.12245 

.99248 

•13975 

.99019 

.15701 

.98760 

58 

3 

.08803 

.99612 

•  10540 

•99443 

.12274 

.99244 

.14004 

.99015 

•1573° 

•98755 

57 

4 

.08831 

.99609 

.  10569 

.99440 

.  12302 

.99240 

•14033 

.99011 

•15758 

•98751 

56 

5 

.08860 

.99607 

•  10597 

•99437 

.12331 

•99237 

.  14061 

.99006 

•15787 

.98746 

55 

6 

.08889 

.99604 

.  10626 

•99434 

.  12360 

•99233 

.14090 

.99002 

.15816 

.98741 

54 

7 

.08918 

.99602 

.  10655 

•99431 

•12389 

•99230 

.14119 

.98998 

•15845 

•98737 

53 

8 

.08947 

.99599 

.  10684 

.99428 

.12418 

.99226 

.14148 

.98994 

>  15873 

.98732 

52 

9 

.08976 

•99596 

.10713 

.99424 

•  12447 

.99222 

.14177 

.98990 

.15902 

.98728 

51 

10 

.09005 

•99594 

.  10742 

.99421 

.12476 

.99219 

.  14205 

.98986 

•15931 

.98723 

So 

ii 

.09034 

•99591 

.10771 

.99418 

.  12504 

.99215 

•14234 

.98982 

•15959 

.98718 

49 

12 

.09063 

.99588 

.10800 

•99415 

•12533 

.99211 

.14263 

.98978 

.15988 

.98714 

48 

13 

.09092 

.99586 

.  10829 

.99412 

.12562 

.99208 

.14292 

•98973 

.16017 

.98709 

47 

14 

.09121 

•99583 

.10858 

•99409 

•12591 

.99204 

.14320 

.98969 

.16046 

.98704 

46 

15 

.09150 

•9958o 

.10887 

.99406 

.  12620 

.99200 

•  14349 

.98965 

.16074 

.98700 

45 

16 

.09179 

•99578 

.10916 

.99402 

.12649 

•99197 

•14378 

.98961 

.16103 

•98695 

44 

17 

.09208 

•99575 

•  10945 

•99399 

.12678 

•99193 

.  14407 

•98957 

.16132 

.98690 

43 

18 

•09237 

•99572 

•10973 

•99396 

.  12706 

.99189 

.14436 

•98953 

.  16160 

.98686 

42 

19 

.09266 

•99570 

.11002 

•99393 

•12735 

.99186 

.14464 

.98948 

.16189 

.98681 

41 

20 

.09295 

•99567 

.IIO31 

.99390 

.  12764 

.99182 

•14493 

•98944 

.16218 

.98676 

40 

21 

.09324 

.99564 

.  Ilo6o 

.99386 

•12793 

.99178 

.14522 

.98940 

.16246 

.98671 

39 

22 

•09353 

.99562 

.11089 

•99383 

.12822 

•99175 

•14551 

.98936 

.16275 

.98667 

38 

23 

.09382 

•99559 

,llll8 

.90380 

.12851 

•99171 

.14580 

.98931 

.16304 

.98662 

37 

24 

.09411 

•99556 

.11147 

•99377 

.12880 

•99167 

.  14608 

.98927 

•  16333 

•98657 

36 

25 

.09440 

•99553 

.11176 

•99374 

.12908 

•99163 

•14637 

.98923 

.16361 

.98652 

35 

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.09469 

•99551 

.  II2O5 

•9937° 

•12937 

.99160 

.14666 

.98919 

.16390 

.98648 

34 

27 

.09498 

.99548 

.11234 

•99367 

•99156 

.14695 

.98914 

.16419 

.98643 

33 

28 

•09527 

•99545 

.11263 

.99364 

•12995 

•99152 

•14723 

.98910 

'.  16447 

.98638 

32 

29 

•09556 

•99542 

.11291 

.99360 

.13024 

.99148 

•14752 

.98906 

.16476 

•98633 

30 

•09585 

.99540 

.11320 

•99357 

•  13053 

•99144 

.14781 

.98902 

.  16505 

.98629 

3° 

31 

.09614 

•99537 

•  "349 

•99354 

.13081 

.99141 

.14810 

.98897 

•16533 

.98624 

29 

32 

.09642 

•99534 

.11378 

•99351 

.13110 

•99137 

.  14838 

.98893 

.16562 

.98619 

28 

33 

.09671 

•99531 

.11407 

•99347 

•13139 

•99133 

.  14867 

.98889 

.16591 

.98614 

27 

34 

.09700 

.99528 

.11436 

•99344 

.13168 

.99129 

.14896 

.98884 

•.16620 

.98609 

26 

35 

.09729 

.99526 

.11465 

•99341 

•13197 

•99125 

.14925 

.98880 

.16648 

.98604 

25 

36 

.09758 

•99523 

.11494 

•99337 

.  13226 

.99122 

•14954 

.98876 

.16677 

.98600 

24 

37 

•09787 

.99520 

•"523 

•99334 

•13254 

.99118 

.14982 

.98871 

.16706 

•98595 

23 

38 

.09816 

•99517 

•"552 

•99331 

•13283 

.99114 

.15011 

.98867 

•16734 

.98590 

22 

39 

•09845 

•99514 

.11580 

.99327 

•13312 

.99110 

.15040 

.98863 

.16763 

.98585 

21 

40 

.09874 

•995" 

.11609 

.99324 

•13341 

.99106 

.15069 

.98858 

.16792 

.98580 

20 

4i 

.09903 

.99508 

.11638 

.99320 

.1337° 

.99102 

•15097 

.98854 

.16820 

•98575 

*9 

42 

.09932 

.99506 

.11667 

•99317 

•13399 

.99098 

.15126 

.98849 

.  16849 

.98570 

18 

43 

.09961 

•99503 

.11696 

•99314 

•13427 

•99094 

•15155 

.98845 

.16878 

•98565 

17 

44 

.09990 

•99500 

.11725 

.99310 

•13456 

.99091 

.15184 

.98841 

.16906 

•98561 

l6 

45 

.  10019 

•99497 

•"754 

•99307 

•13485 

.99087 

.  15212 

.98836 

•16935 

•98556 

15 

46 

.  10048 

•99494 

.  11783 

•993°3 

•13514 

.99083 

.15241 

.98832 

.16964 

•98551 

M 

47 

.10077 

.99491 

.11812 

.99300 

•13543 

•99°79 

.15270 

.98827 

.16992 

•98546 

48 

.10106 

.99488 

.11840 

•99297 

•13572 

•99075 

•15299 

.98823 

.17021 

.98541 

12 

49 
.5° 

.  10135 
.  10164 

.99485 
.99482 

.  i  i  869 
.11898 

•99293 
.99290 

.  13600 
.  13629 

.99071 
.99067 

•15327 
•15356 

.98818 
.98814 

.17050 
.17078 

•98536 
•98531 

11 
IO 

5I 

.10192 

•99479 

.11927 

.99286 

.13658 

.99063 

.15385 

.98809 

.17107 

•98526 

9 

S2 

.  IO22I 

.99476 

.11956 

.99283 

•13687 

.99059 

•15414 

.98805 

•17136 

.98521 

8 

53 

.  10250 

•99473 

.11985 

•99279 

.13716 

•99055 

.15442 

.  98800 

.17164 

•98516 

7 

54 

.  10279 

.99470 

.12014 

.99276 

.13744 

.99051 

•15471 

.98796 

•17193 

.985" 

6 

.10308 

•99467 

.12043 

.99272 

•13773 

.99047 

•15500 

.98791 

.17222 

.98506 

5 

56 

10337 

.99464 

.12071 

.99269 

.13802 

.99043 

•15529 

.98787 

.17250 

.98501 

4 

57 

.  10366 

.99461 

.I2IOO 

.99265 

•13831 

•99039 

•15557 

.98782 

•17279 

.98496 

3 

58 

•  10395 

•99458 

.  I2I29 

.99262 

.13860 

•99°35 

•15586 

.98778 

.17308 

.98491 

2 

59 

.  10424 

•99455 

.12158 

.99258 

.13889 

•99031 

•15615 

•98773 

•17336 

.98486 

I 

60 

•  1°453 

•99452 

.12187 

.99255 

•13917 

.99027 

•15643 

.98769 

•17365 

.98481 

o 

Cosine 

Sine 

Cosine 

Sine 

Cosine 

Sine 

Cosine 

Sine 

Cosine 

Sine 

i 

' 

84° 

83° 

82° 

81° 

80° 

NATURAL  SINES  AND  COSINES. 


79 


10° 

i     H° 

12° 

13° 

14° 

, 

Sine 

Cosine 

Sine 

Cosine 

Sine 

Cosine 

Sine 

Cosine 

Sine 

Cosine 

0 

•17365 

.98481 

.19081 

•98163 

.20791 

•97815 

.22495 

•97437 

.24192 

.97030 

60 

I 

•17393 

.98476 

.19109 

.98157 

.20820 

.97809 

•22523 

•9743° 

,  24220 

.97023 

59 

2 

.17422 

.98471 

.19138 

.98152 

.20848 

•97803 

•22552 

.97424 

.24249 

•97015 

58 

3 

•17451 

.98466 

.19167 

.98146 

.20877 

•97797 

.22580 

•97417 

•24277 

.97008 

57 

4 

•17479 

.98461 

•19195 

.98140 

.20905 

•97791 

.22608 

.97411 

•24305 

.97001 

56 

5 

.17508 

•98455 

.19224 

•98135 

•20933 

.97784 

•22637 

.97404 

•24333 

.96994 

55 

6 

•17537 

.98450 

.19252 

'  .98129 

.  20962 

•97778 

.22665 

.97398 

.24362 

.96987 

54 

7 

•17565 

.98445 

.  19281 

.98124 

.20990 

•97772 

.22693 

•97391 

.24390 

.96980 

53 

8 

•17594 

.98440 

.19309 

.98118 

.21019 

•97766 

.22722 

•97384 

.24418 

•96973 

52 

9 

.17623 

•98435 

.19338 

.98112 

.21047 

•9776o 

.22750 

•97378 

.  24446 

.96966 

51 

10 

.17651 

.98430 

.19366 

.98107 

.21076 

•97754 

.22778 

•9737  ! 

.24474 

•96959 

5° 

ii 

.17680 

.98425 

•19395 

.98101 

.21104 

.97748 

.22807 

•97365 

•24503 

.96952 

4Q 

12 

.17708 

.98420 

•  19423 

.98096 

.21132 

•97742 

.22835 

•97358 

•24531 

.96945 

48 

13 

•  17737 

.98414 

.19452 

.98090 

.21161 

•97735 

.22863 

•97351 

•24559 

.96937 

47 

14 

.  17766 

.98409 

.19481 

.98084 

.21189 

•97729 

.22892 

•97345 

•24587 

.96930 

46 

15 

•17794 

.98404 

•  19509 

.98079 

.21218 

•97723 

.22920 

•97338 

.24615 

.96923 

45 

16 

•17823 

.98399 

•19538 

.98073 

.21246 

•97717 

.22948 

•97331 

.24644 

.96916 

44 

17 

•17852 

.98394 

•  19566 

.98067 

.21275 

.97711 

.22977 

•97325 

.24672 

.96909 

43 

18 

.17880 

.98389 

•  19595 

.98061 

.21303 

•97705 

•23005 

•97318 

.24700 

.96902 

42 

19 

.17909 

•98383 

•19623 

.98056 

•21331 

.97698 

•23033 

•973" 

.24728 

.96894 

20 

•  17937 

.98378 

.19652 

.98050 

.21360 

.97692 

.23062 

•97304 

•24756 

.96887 

40 

21 

.17966 

•98373 

.  19680 

.98044 

.21388 

.97686 

.23090 

.97298 

•24784 

.96880 

39 

22 

•17995 

.98368 

.19709 

.98039 

.21417 

.97680 

.23118 

•97291 

.24813 

•96873 

38 

23 

.18023 

.98362 

•19737 

,98033 

.21445 

•97673 

•23146 

.97284 

.24841 

.96866 

37 

24 

.  18052 

•98357 

.19766 

.98027 

.21474 

.97667 

•23175 

.97278 

.24869 

.96858 

36 

25 

.18081 

•98352 

•  19794 

.98021 

.21502 

.97661 

•23203 

•97271 

.24897 

.96851 

35 

26 

.  18109 

•98347 

•  19823 

.98016 

.21530 

•97655 

.23231 

•97264 

•24925 

.96844 

34 

27 

.  18138 

.98341 

.19851 

.98010 

•21559 

.97648 

.23260 

•97257 

•24954 

,96837 

33 

28 

.18166 

•98336 

.19880 

.98004 

.21587 

•97642 

.23288 

•97251 

.24982 

.96829 

32 

29 

•18195 

•98331 

.  19908 

.97998 

.21616 

.97636 

.23316 

.97244 

.25010 

.96822 

31 

30 

.18224 

•98325 

•19937 

.97992 

.21644 

.97630 

•23345 

•97237 

•25038 

.96815 

30 

31 

.18252 

.98320 

•  !9965 

.97987 

.21672 

.97623 

•23373 

.97230 

.  25066 

.96807 

29 

32 

.  £828  1 

•98315 

•  19994 

.97981 

.21701 

.97617 

.23401 

.97223 

.25094 

.96800 

28 

33 

.18309 

•98310 

.20022 

•97975 

.21729 

.97611 

•23429 

.97217 

.25122 

.96793 

27 

34 

•18338 

.98304 

.20051 

•97969 

.21758 

.97604 

•23458 

.97210 

•25151 

.96786 

26 

35 

.18367 

.98299 

.20079 

•97963 

.21786 

•97598 

.23486 

•97203 

•25179 

.96778 

25 

36 

•18395 

.98294 

.20108 

•97958 

.21814 

•97592 

•23514 

.97196 

.25207 

.96771 

24 

37 

.18424 

.98288 

.20136 

•97952 

.21843 

•97585 

•23542 

.97189 

•25235 

.96764 

23 

38 

.18452 

.98283 

.20165 

.97946 

.21871 

•97579 

•23571 

.97182 

.25263 

.96756 

22 

39 

.18481 

.98277 

.20193 

.97940 

.21899 

•97573 

•23599 

.97176 

•25291 

.96749 

21 

40 

.18509 

.98272 

.20222 

•97934 

.21928 

.97566 

•23627 

.97169 

.25320 

•96742 

2O 

4i 

•18538 

.98267 

.20250 

.97928 

.21956 

.97560 

.23656 

.97162 

•25348 

.96734 

Ig 

42 

.18567 

.98261 

.20279 

.97922 

.21985 

•97553 

.23684 

•97155 

•25376 

.96727 

18 

43 

•18595 

.98256 

.20307 

•979i6 

.22013 

•97547 

.23712 

.97148 

.25404 

•96719 

17 

44 

.18624 

.98250 

•20336 

.97910 

.22041 

•97541 

•23740 

.97141 

•25432 

.96712 

16 

45 

.  18652 

.98245 

.20364 

•97905 

.  22070 

•97534 

•23769 

•97134 

.25460 

.  96705 

15 

46 

.  18681 

.98240 

-.20393 

.97899 

.22098 

.97528 

•23797 

.97127 

.25488 

.96697 

14 

47 

.18710 

.98234 

.20421 

•97893 

.22126 

•97521 

.23825 

.97120 

•25516 

.96690 

48 

.18738 

.98229 

.20450 

.97887 

.22155 

•97515 

•23853 

•97"3 

•25545 

.96682 

12 

49 

.18767 

.98223 

.20478 

.97881 

.22183 

.97508 

.23882 

.97106 

•25573 

.96675 

II 

5° 

•18795 

.98218 

.20507 

•97875 

.22212 

•97502 

.23910 

.97100 

.25601 

,96667 

IO 

Si 

.  18824 

.98212 

•20535 

.97869 

.2224O 

•97496 

•23938 

.97093 

•25629 

.96660 

9 

S2 

.18852 

.98207 

•20563 

.97863 

.22268 

.97489 

.23966 

.97086 

•25657 

.96653 

8 

53 

.18881 

.98201 

.  20592 

•97857 

.22297 

.97483 

•23995 

•97079 

•25685 

.96645 

7 

54 

.18910 

.98196 

.  20620 

•97851 

•22325 

.97476 

.24023 

.97072 

•25713 

.96638 

6 

55 

.18938 

.98190 

.20649 

•97845 

•22353 

.97470 

.24051 

•97065 

•25741 

.96630 

5 

56 

.18967 

.98185 

.20677 

•97839 

.22382 

•97463 

.24079 

.97058 

.25769 

.96623 

4 

57 

.18995 

.98179 

.  20706 

•97833 

.224IO 

•97457 

.24108 

•97051 

.25798 

.96615 

3 

58 

.19024 

.98174 

•20734 

.97827 

•22438 

•9745° 

.24136 

•97044 

.25826 

.96608 

2 

59 

.19052 

.98168 

.20763 

.97821 

.22467 

•97444 

.24164 

.97037 

•25854 

.96600 

I 

60 

.19081 

.98163 

.20791 

•97815 

.22495 

•97437 

.24192 

.97030 

.25882 

•96593 

o 

, 

Cosine 

Sine 

Cosine 

Sine 

Cosine 

Sine 

Cosine 

Sine 

Cosine 

Sine 

, 

79° 

78° 

77° 

76° 

75° 

80 


NATURAL  SINES  AND  COSINES. 


'5° 

16° 

i?° 

18° 

19° 

Sine 

Cosine 

Sine 

Cosine 

Sine 

Cosine 

Sine 

Cosine 

Sine 

Cosine 

o 

I 

.25882 
.25910 

•96593 
•96585 

.27564 
.27592 

.96126 
.96118 

.29237 
.29265 

•9563° 

.95622 

.30902 
•30929 

.95106 
•95097 

•32557 
•32584 

•94552 
•94542 

60 
59 

2 

•25938 

.96578 

.27620 

.96110 

.29293 

•95613 

•  30957 

.95088 

.32612 

•94533 

58 

3 

.25966 

•9657° 

.27648 

.96102 

.29321 

•95605 

•30985 

•95°79 

.32639 

•94523 

57 

4 

•25994 

.96562 

.27676 

.96094 

.29348 

•95596 

.31012 

.95070 

.32667 

•945M 

56 

5 

.26022 

•96555 

.27704 

.96086 

.29376 

•95588 

.31040 

.95061 

•  32694 

•94504 

5S 

6 

.  26050 

•96547 

•27731 

.96078 

.  29404 

•95579 

.31068 

•95052 

•32722 

•94495 

54 

7 

.26079 

.96540 

•27759 

.96070 

•29432 

•95571 

•31095 

•95°43 

•32749 

.94485 

53 

8 

.26107 

•96532 

.27787 

.96062 

.  29460 

•95562 

•3"23 

•95033 

•32777 

.94476 

52 

9 

.26135 

.96524 

.27815 

.96054 

.29487 

•95554 

•3II5I 

•95024 

.32804 

.94466 

5i 

10 

.26163 

•965:7 

•27843 

.96046 

•295*5 

•95545 

.31178 

•95015 

•32832 

•94457 

50. 

ii 

.26191 

.96509 

.27871 

.96037 

•29543 

•95536 

.31206 

.95006 

•32859 

•94447 

49 

12 

.26219 

.96502 

.27899 

.96029 

•29571 

•95528 

•31233 

•94997 

.32867 

.94438 

48 

*3 

.26247 

.96494 

.27927 

.96021 

.29599 

•955*<) 

.31261 

.94988 

•329H 

.94428 

47 

M 

.26275 

.96486 

•27955 

•96013 

.29626 

•955" 

.31289 

•94979 

•32942 

.94418 

46 

•26303 

.96479 

•27983 

.96005 

.29654 

•95502 

•31316 

.94970 

.32969 

.94409 

45 

16 

•26331 

.96471 

.28011 

•95997 

.29682 

•95493 

•3*344 

.94961 

.32997 

•94399 

44 

i? 

•26359 

.96463 

•28039 

.95989 

.29710 

•95485 

•3T372 

.94952 

•  33024 

.94390 

43 

18 

.26387 

•96456 

.28067 

.95981 

•29737 

•95476 

•31399 

•94943 

•33051 

.94380 

42 

T9 

.26415 

.96448 

•  28095 

•95972 

•29765 

•95467 

•3M27 

•94933 

•33079 

•9437° 

4i 

20 

•26443 

.96440 

.28123 

.95964 

.29793 

•95459 

•3T454 

.94924 

.33106 

.94361 

40 

21 

.26471 

•96433 

.28150 

•95956 

.29821 

•95450 

.31482 

•949!5 

•33134 

•94351 

39 

22 

.26500 

.96425 

.28178 

.95948 

.29849 

•95441 

.31510 

.94906 

•33l6i 

.94342 

38 

23 

.26528 

.96417 

.28206 

.95940 

.29876 

•95433 

•31537 

.94897 

•33189 

•94332 

37 

24 

.26556 

.96410 

.28234 

•95931 

.29904 

.95424 

•3I565 

.94888 

.33216 

•94322 

36 

25 

.26584 

.96402 

.28262 

•95923 

.29932 

•95415 

•3*593 

.94878 

•33244 

•94313 

35 

26 

.26612 

.  963-^4 

.28290 

•959!5 

.  29960 

•954°7 

.31620 

.94869 

•33271 

•943°3 

34 

27 

.  26640 

.96386 

.28318 

•95907 

.29987 

•95398 

.31648 

.94860 

•33298 

.94293 

33 

28 

.26668 

•96379 

.28346 

.95898 

•30015 

•95389 

•3l675 

.94851 

.33326 

.94284 

32 

29 

.26696 

•96371 

•28374 

.95890 

.30043 

•9538o 

•31703 

.94842. 

•33353 

•94274 

3i 

3° 

.26724 

.96363 

.  28402 

.95882 

.30071 

•95372 

•3173° 

.94832 

•333Sl 

.94264 

3° 

31 

•26752 

•96355 

.28429 

•95874 

.30098 

•95363 

.31758 

•94823 

.33408 

.94254 

29 

S2 

.26780 

•96347 

•28457 

•95865 

.30126 

•95354 

.31786 

.94814 

.33436 

•94245 

28 

33 

.26808 

.96340 

.28485 

•95857 

•3OI54 

•95345 

•31813 

.94805 

•33463 

.94235 

27 

34 

.26836 

.96332 

•28513 

•95849 

.30182 

•95337 

.31841 

•94795 

•  3349° 

.94225 

26 

35 

.26864 

.96324 

.28541 

.95841 

.30209 

•95328 

.31868 

.94786 

.335i8 

.94215 

25 

36 

.26892 

.96316 

.28569 

.95832 

•30237 

•953*9 

.31896 

•94777 

•33545 

.94206 

24 

37 

.26923 

.96308 

•28597 

.95824 

.30265 

•953!o 

•3I923 

.94768 

•33573 

.94196 

23 

38 

.26948 

.96301 

.28625 

.95816 

.30292 

•95301 

•3*95* 

•94758 

.33600 

.94186 

22 

39 

.26976 

.96293 

.28652 

.95807 

.30320 

•95203 

•3*979 

•94749 

•33627 

.94176 

21 

40 

.27004 

.96285 

.28680 

•95799 

•30348 

•95284 

.32006 

.94740 

•33655 

.94167 

2O 

41 

.27032 

.96277 

.28708 

•95791 

•30376 

•95275 

•32034 

•9473° 

.33682 

•94I57 

19 

42 

.27060 

.96269 

.28736 

.95782 

-30403 

.95266 

.32061 

.94721 

•33710 

.94147 

18 

43 

.27088 

.96261 

.28764 

•95774 

•30431 

•95257 

.32089 

.94712 

•33737 

•94137 

17 

44 

.27116 

.96253 

.28792 

.95766 

•  30459 

.95248 

.32116 

.94702 

•33764 

.94127 

16 

45 

.27144 

.96246 

.28820 

•95757 

.30486 

.95240 

.32144 

•94693 

•33792 

.94118 

15 

46 

.27172 

•96238 

.28847 

•95749 

•  3°5  T4 

•95231 

.32171 

.94684 

•33819 

.94108 

14 

47 

.27200 

.96230 

.28875 

•95740 

•  30542 

•95222 

.32199 

.94674 

•33846 

.94098 

*3 

48 

.27228 

.96222 

.28903 

•95732 

•30570 

•95213 

.32227 

.94665 

•33874 

.94088 

12 

49 

.27256 

.96214 

.28931 

•95724 

•  3°597 

.95204 

•32254 

.94656 

•33901 

.94078 

11 

50 

.27284 

.96206 

.28959 

•95715 

.30625 

-95!95 

.32282 

.94646 

•33929 

.94068 

10 

5* 

.27312 

.96198 

.28987 

•95707 

•30653 

.95186 

.32309 

•94637 

•33956 

.94058 

9 

52 

•27340 

.96190 

.29015 

.95698 

.30680 

•95!77 

•32337 

.94627 

•33983 

.94049 

8 

53 

•27368 

.96182 

.29042 

.95690 

.30708 

.95168 

•32364 

.94618 

.34011 

•94039 

7 

54 

.27396 

.96174 

.29070 

.95681 

.30736 

•95^9 

.32392 

.94609 

•34038 

.  94029 

6 

55 

.27424 

.96166 

.29098 

•95673 

•30763 

•95150 

•32419 

•94599 

•34065 

.94019 

5 

56 

•27452 

.96158 

.29126 

.95664 

.30791 

.95142 

•32447 

•94590 

•34093 

.94009 

4 

57 

.27480 

.96150 

.29154 

•95656 

.30819 

•95*33 

•32474 

•9458o 

.34120 

•93999 

3 

58 

.27508 

.96142 

.29182 

•95647 

.30846 

•95124 

.32502 

•94571 

•34J47 

.93989 

2 

59 

.27536 

.96134 

.29209 

•95639 

.30874 

•95"5 

•32529 

.94561 

•34175 

•93979 

I 

60 

•27564 

.96126 

.29237 

.95630 

.  30002 

.95106 

•32557 

•94552 

.34202 

•93969 

0 

Cosine 

Sine 

Cosine 

Sine 

Cosine 

Sine 

Cosine 

Sine 

Cosine 

Sine 

r 

1 

74°       73° 

72° 

71° 

7°° 

NATURAL  SINES  AND  COSINES. 


81 


2 

0° 

2 

. 

2 

o 
2 

2 

3°    i 

2. 

1° 

Sine 

Cosine 

Sine 

Cosine 

Sine 

Cosine 

Sine 

Cosine 

Sine 

Cosine 

o 

.34202 
.34229 

•93969 
•93959 

•35837 
.35864 

•93358 
•93348 

•3746* 
.37488 

.92718 
.92707 

•39073 
.39100 

.92050 
.92039 

.40674 
.40700 

•9*355 
•9*343 

60 
59 

2 

•34257 

•93949 

•3589* 

•93337 

•375'5 

.92697 

•39*27 

.92028 

.40727 

•9*33* 

58 

3 

•34284 

•93939 

.35918 

•93327 

•37542 

.92686 

•39*53 

.92016 

•40753 

•9*3*9 

57 

4 

•343** 

•93929 

•35945 

•933*6 

•37569 

.92675 

•39*83 

.  92005 

.40780 

•9*307 

56 

5 

•34339 

.93919 

•35973. 

•93306 

•37595 

.92664 

•39207 

.91994 

.40806 

.91295 

55 

6 

.34366 

•93909 

.36000 

•93295 

•37622 

•92653 

•39234 

.91982 

.40833 

.91283 

54 

7 

•  34393 

•93899 

.36027 

•93285 

•37649 

.92642 

.39260 

.91971 

.40860 

.91272 

53 

8 

•3442* 

.93889 

•36054 

•93274 

•37676 

•92631 

•39287 

•9*959 

.40886 

.91260 

52 

9 

.34448 

.93879 

.36081 

.93264 

•37703 

.92620 

.39314 

.91948 

.40913 

.91248 

5* 

10 

•34475 

.93869 

.36108 

•93253 

.37730 

.92609 

•3934* 

•9*936 

.40939 

.91236 

50 

ii 

•34503 

•93859 

•36*35 

•93243 

•37757 

•92598 

•39367 

.91925 

.40966 

.91224 

49 

12 

••34530 

•93849 

.36162 

•93232 

•37784 

.92587 

•39394 

.91914 

.40992 

.91212 

48 

*3 

•34557 

•93839 

.36190 

.93222 

.37811 

•92576 

•3942* 

.91902 

.41019 

.91200 

47 

•34584 

.93829 

.36217 

.93211 

•37838 

•92565 

•3944s 

.91891 

.41045 

.91188 

46 

15 

•346*2 

.93819 

•36244 

•93201 

•37865 

•92554 

•  39474 

•9*879 

.41072 

.91176 

45 

16 

•  34639 

.93809 

.36271 

.93190 

.37892 

•92543 

•3950* 

.91868 

.41098 

.91164 

44 

*7 

.34666 

•93799 

.3-6298 

•93*80 

•379*9 

•92532 

•39528 

.91856 

.  1125 

.91152 

43 

18 

•34694 

.93789 

•36325 

•93*69 

•37946 

.92521 

•39555 

.91845 

•  **5* 

.91140 

42 

*9 

.34721 

•93779 

•93*59 

•37973 

.92510 

•3958* 

•9*833 

.  1178 

.91128 

20 

.34748 

•93769 

•36379 

•93*48 

•37999 

•92499 

.39608 

.91822 

.  1204 

.91116 

40 

21 

•34775 

•93759 

.36406 

•93*37 

.38026 

.92488 

•39635 

.91810 

•  123* 

.91104 

39 

22 

•34803 

•93748 

•36434 

•93*27 

•38053 

.92477 

.39661 

•9*799 

.91092 

38 

23 

•34830 

•9373s 

.36461 

.93116 

.38080 

.92466* 

.39688 

.91787 

.  1284 

.91080 

37 

24 

.34857 

.93728 

.36488 

.93106 

.38107 

•92455 

•397*5 

•9*775 

.  1310 

.91068 

36 

25 

.34884 

.93718 

•365*5 

•93095 

•38*34 

.92444 

•3974* 

•9*764 

•  *337 

.91056 

35 

26 

•349*2 

•93708 

.36542 

•93084 

.38161 

.92432 

•39768 

•9*752 

•  *363 

•9*°44 

34 

27 

•34939 

.93698 

.36569 

•93°74 

.38188 

.92421 

•39795 

.91741 

•  *39° 

.91032 

33 

28 

•34966 

.93688 

.36596 

.93063 

•382:5 

.92410 

.39822 

•9*729 

.  1416 

.91020 

S2 

29 

•  34993 

•93677 

.36623 

•93052 

.38241 

•92399 

.39848 

.91718 

•4*443 

.91008 

31 

3° 

.35021 

•93667 

.36650 

•93042 

.  38268 

.92388 

•39875 

.91706 

.41469 

.90996 

3° 

3i 

•35048 

•93657 

.36677 

•93°3* 

•38295 

•92377 

.39902 

.91694 

.41496 

.90984 

29 

32 

•35075 

.93647 

•36704 

.93020 

'38322 

.92366 

•39928 

•9*683 

•4*522 

.90972 

28 

33 

•35*02 

•93637 

•3673* 

.93010 

•38349 

•92355 

•39955 

.91671 

•4*549 

.90960 

27 

34 

•35*30 
•35*57 

.93626 
.93616 

•36758 
•36785 

•92999 
.92988 

•38376 
.38403 

•92343 
.92332 

.39982 
.40008 

.91660 
.91648 

•4*575 
.41602 

.90948 
.90936 

26 
25 

36 

•35*84 

.93606 

.36812 

.92978 

.38430 

.92321 

.40035 

.91636 

.41628 

.90924 

24 

37 

•352*1 

•93596 

•36839 

.92967 

•38456 

•923*0 

.40062 

.91625 

•4*655 

23 

38 

•35239 

•93585 

.36867 

•92956 

.38483 

.92299 

.40088 

.91613 

.41681 

.90899 

22 

39 

.35266 

•93575 

.36894 

•92945 

.38510 

.92287 

.40115 

.91601 

•4*707 

.90887 

21 

40 

•35293 

•93565 

.36921 

•92935 

.38537 

.92276 

.40141 

.91590 

•4*734 

•90&75 

20 

4i 

•35320 

•93555 

.36948 

.92924 

.38564 

.92265 

.40168 

•9*578 

.41760 

.90863 

*9 

42 

•35347 

•93544 

•36975 

•929*3 

.3859* 

.92254 

.40195 

.9*566 

.41787 

•90851 

18 

43 

•35375 

•93534 

.37002 

.  92902 

.38617 

•92243 

.40221 

•9*555 

.41813 

.90839 

17 

44 

•35402 

•93524 

.37029 

.92892 

•38644 

.92231 

.40248 

•9*543 

.  1840 

.90826 

16 

45 

•35429 

•935*4. 

•37056 

.92881 

-3867* 

.92220 

.40275 

•9*53* 

.  1866 

.90814 

IS 

46 

•35456 

.•93503 

•37083 

.92870 

.38698 

.92209 

.40301 

•9*5*9 

.  1892 

.90802 

14 

47 

•35484 

•93493 

.37110 

.92859 

•38725 

.92198 

.40328 

.91508 

•  *9*9 

.90790 

*3 

48 

•355** 

•93483 

•37*37 

.92849 

•38752 

.92186 

•40355 

.91496 

•  *945 

.90778 

12 

49 

•35538 

.93472 

.37164 

.92838 

.38778 

•92*75 

.40381 

.91484 

•  1972 

.90766 

II 

50 

•35565 

.93462 

•37*9* 

.92827 

.38805 

.92164 

.40408 

•9*472 

.  1998- 

•9°753 

IO 

5I 

•35592 

•93452 

.37218 

.92816 

.38832 

•92*52 

•40434 

.91461 

.42024 

.90741 

9 

52 

.35619 

.93441 

•37245 

.92805 

•38859 

.92141 

.40461 

•9*449 

.42051 

.90729 

8 

53 

•35647 

•9343* 

.37272 

.02794 

.38886 

.92130 

.40488 

•9*437 

.42677 

.90717 

7 

54 

•35674 

.93420 

•37299 

.92784 

•389*2 

.92119 

.40514 

•9*425 

.42104 

.  90704 

6 

55 

•357°* 

•934*0 

•37326 

.92773 

•38939 

.92107 

.40541 

•9*4*4 

.42*3° 

.90692 

5 

56 

•35728 

•934oo 

•37353 

.92762 

.38966 

.  92096 

.40567 

.91402 

.42156 

.90680 

4 

57 

•35755 

•93389 

•3738o 

•9275* 

•38993 

.92085 

.  -4°594 

.91390 

.42183 

.90668 

3 

58 

•35782 

•93379 

•  37407 

.92740 

.39020 

•92073 

.40621 

•9*378 

.42209 

•90655 

2 

59 
60 

.358*0 
.35837 

•93368 
•93358 

•37434 
•3746* 

.92729 
.92718 

.39046 
•39073 

.92062 
.92050 

.40647 
.40674 

.91366 
•9*355 

•42235 
.42262 

.90643 
.  9063  i 

I 
O 

, 

Cosine 

Sine 

Cosine 

Sine 

Cosine 

Sine 

Cosine 

Sine 

Cosine 

Sine 

, 

6 

9 

6 

3° 

6 

7° 

6 

6° 

6 

5° 

NATURAL  SINES  AND  COSINES. 


'5° 

26° 

27° 

28° 

29° 

Sine 

Cosine 

Sine 

Cosine 

Sine 

Cosine 

Sine 

Cosine 

Sine 

Cosine 

0 

I 

.42262 
.42288 

.90631 
.90618 

•43837 
•43863 

.89879 
.89867 

•45399 
•45425 

.89101 
.89087 

.46947 
•46973 

.88295 
.88281 

.48481 
.48506 

.87462 
.87448 

60 
59 

2 

•423*5 

.90606 

.43889 

•89854 

•4545* 

.89074 

.46999 

.88267 

.48532 

•87434 

58 

3 

.42341 

.90594 

•439*6 

.89841 

•45477 

.89061 

•47024 

.88254 

•48557 

.87420 

57 

4 

•42367 

.90582 

.43942 

.  89828 

.45503 

.  89048 

.47050 

.  88240 

.48583 

.87406 

56 

5 

•42394 

.90569 

.43968 

.89816 

.45529 

•89035 

.47076 

.88226 

.48608 

•8739* 

55 

6 

.42420 

•90557 

•  43994 

.89803 

•45554 

.89021 

.47101 

.88213 

.48634 

•87377 

54 

7 

.42446 

•90545 

.44020 

.89790 

•4558o 

.89008 

•47*27 

.88199 

.48659 

•87363 

53 

8 

•42473 

•90532 

.44046 

.89777 

.45606 

.88995 

•47*53 

.88185 

.48684 

•87349 

52 

9 

•42499 

.90520 

.44072 

.89764 

•45632 

.  88o8l 

•47*78 

.88172 

.48710 

•87335 

5* 

10 

•42525 

.90507 

.44098 

•89752 

•45658 

.88968 

.47204 

.88158 

•48735 

.87321 

50 

ii 

•42552 

.90495 

•44*24 

•89739 

.45684 

•88955 

•47229 

.88144 

.48761 

.87306 

49 

12 

•42578 

.90483 

•44*5* 

.89726 

•457*0 

.88942 

•47255 

.88130 

.48786 

.87292 

48 

*3 

.42604 

.90470 

•44*77 

•897*3 

•45736 

.88928 

.47281 

.88117 

.48811 

.87278 

47 

.42631 

.90458 

.44203 

.89700 

.45762 

.88915 

.47306 

.88103 

.48837 

.87264 

46 

15 

•42657 

.  90446 

.44229 

.89687 

.45787 

.88902 

•47332 

.88089 

.48862 

.87250 

45 

16 

.42683 

•90433 

•44255 

.89674 

•458*3 

.88888 

•47358 

.88075 

.48888 

•87235 

44 

*7 

.42709 

.90421 

.44281 

.89662 

•45839 

•88875 

•47383 

.88062 

.48913 

.87221 

43 

18 

.42736 
.42762 

.  90408 
•90396 

•44307 
•44333 

.89649 
.89636 

.45865 
.45891 

.88862 
.88848 

•47409 
•47434 

.88048 
.88034 

.48938 
.48964 

.87207 
•87193 

42 

20 

.42788 

.90383 

•44359 

.89623 

•459*7 

.88835 

.47460 

.88020 

.48989 

.87178 

40 

21 

.42815 

•9037* 

•44385 

.89610 

•45942 

.88822 

.47486 

.88006 

.49014 

.87164 

39 

22 

.42841 

.90358 

.44411 

•89597 

.45968 

.88808 

•475** 

•87993 

.49040 

.87150 

38 

23 

.42867 

•90346 

•44437 

•89584 

•45994 

.88795 

•47537 

.87979 

.49065 

.87136 

37 

24 

.42894 

•90334 

•44464 

•8957* 

.  46020 

.88782 

.47562 

.87965 

.49090 

.87121 

36 

25 

.42920 

.90321 

.44490 

•89558 

.46046 

.88768 

•47588 

•8795* 

.49116 

.87107 

35 

26 

.42946 

.90309 

•445*6 

•89545 

.46072 

•88755 

.47614 

•87937 

.49141 

.87093 

34 

27 

•42972 

.90296 

•44542 

.89532 

.46097 

.88741 

•47639 

.87923 

.49166 

•87079 

33 

28 

•42999 

.90284 

.44568 

.89519 

.46123 

.88728 

•47665 

.87909 

.49192 

.87064 

32 

29 

•43025 

.90271 

•44594 

.89506 

.46149 

•88715 

.47690 

.87896 

.49217 

.87050 

31 

30 

•43051 

.90259 

.44620 

•89493 

•46175 

.88701 

•477*6 

.87882 

.49242 

.87036 

30 

3* 

•43°77 

.90246 

.44646 

.89480 

.46201 

.88688 

•4774* 

.87868 

.49268 

.87021 

29 

32 

.43104 

•90233 

.44672 

.89467 

.46226 

.88674 

•47767 

.87854 

•49293 

.87007 

28 

33 

•43*3° 

.90221 

.44698 

.89454 

.46252 

.88661 

•47793 

.87840 

.49318 

.86993 

27 

34 

.43*56 

.  90208 

•44724 

.89441 

.46278 

.88647 

.47818 

.87826 

•49344 

.86978 

26 

35 

.43*82 

.90196 

•4475° 

.89428 

.46304 

.88634 

•47844 

.87812 

.49369 

.86964 

25 

36 

•43209 

.90183 

•44776 

•894*5 

•4633° 

.88620 

.47869 

.87798 

•49394 

.86949 

24 

37 

•43235 

.90171 

.44802 

.89402 

•46355 

.88607 

•47895 

.87784 

.49419 

•86935 

23 

38 

•43261 

.90158 

.44828 

•89389 

.46381 

•88593 

.47920 

.87770 

•49445 

.86921 

22 

39 

.43287 

.90146 

.44854 

.89376 

.46407 

.88580 

.47946 

.87756 

.49470 

.86906 

21 

40 

•433*3 

•90*33 

.44880 

.89363 

•46433 

.88566 

•4797* 

•87743 

•49495 

.86892 

20 

41 

•43340 

.90120 

.44906 

.89350 

.46458 

•88553 

•47997 

.87729 

•4952* 

.86878 

*9 

42 

•43366 

.90108 

•44932 

•89337 

.  46484 

•88539 

.48022 

•87715 

.49546 

.86863 

18 

43 

•43392 

.90095 

.44958 

•89324 

•465*0 

.88526 

.48048 

.87701 

•4957* 

.86849 

*7 

44 

.43418 

.90082 

•44984 

.89311 

.46536 

.88512 

.48073 

.87687 

•49596 

.86834 

16 

45 

•43445 

.90070 

.45010 

.89298 

.46561- 

.88499 

.48099 

.87673 

.49622 

.86820 

is 

46 

•4347* 

.90057 

•45036 

.89285 

•46587 

.88485 

.48124 

.87659 

.49647 

.86805 

14 

47 

•  43497 

.90045 

.45062 

.89272 

.46613 

.88472 

.48150 

.87645 

.49672 

.86791 

13 

48 

•43523 

.90032 

.45088 

.89259 

.46639 

.88458 

.87631 

.49697 

•86777 

12 

49 

•43549 

.90019 

•45*14 

.89245 

.46664 

•88445 

.48201 

.87617 

•49723 

.86762 

II 

5° 

•43575 

.90007 

•45*40 

.89232 

.46690 

.88431 

.48226 

.87603 

•49748 

.86748 

IO 

5I 

.43602 

•89994 

.45*66 

.89219 

.46716 

.88417 

•48252 

.87589 

•49773 

.86733 

9 

52 

.43628 

.89981 

•45*92 

.89206 

.46742 

.88404 

.48277 

•87575 

.49798 

.86719 

8 

53 

•43654 

.89968 

.45218 

.89193 

.46767 

.88390 

•48303 

.87561 

.49824 

.£6704 

7 

54 

.43680 

.89956 

•45243 

.89180 

•46793 

•88377 

.48328 

.87546 

.49849 

.86690 

6 

55 

.43706 

.89943 

.45269 

.89167 

.46819 

•88363 

•48354 

.87532 

.49874 

.86675 

5 

56 

•43733 

.89930 

•45295 

•89*53 

.46844 

..88349 

•48379 

.87518 

•49899 

.86661 

4 

57 

•43759 

.89918 

•45321 

.89140 

.46870 

.88336 

.48405 

.87504 

•49924 

.86646 

3 

58 

•43785 

•89905 

•45347 

.89127 

.46896 

.88322 

.48430 

.87490 

.49950 

.86632 

2 

59 

.43811 

.89892 

•45373 

.89114 

.46921 

.88308 

.48456 

•87476 

•49975 

.86617 

I 

60 

.89879 

•45399 

.89101 

.46947 

.88295 

.48481 

.87462 

.50000 

.86603 

0 

|, 

Cosine 

Sine 

Cosine 

Sine 

Cosine 

Sine 

Cosine 

Sine 

Cosine 

Sine 

, 

<  . 

64° 

63° 

62° 

6l° 

60° 

•     NATURAL  SINES  AND  COSINES. 


83 


f 

3°°  . 

31° 

32° 

33° 

34° 

Sine 

Cosine 

Sine 

Cosine 

Sine 

Cosine 

Sine 

Cosine 

Sine 

Cosine 

o 

I 

.50000 
.50025 

.86603 
.86588 

•5*529 

•857*7 
.85702 

•52992 
•530*7 

.84805 
.84789 

.54464 
.54488 

•83867 
.8385* 

•559*9 
•55943 

.82904 
.82887 

60 

59 

2 

.50050 

•86573 

.85687 

•5304* 

.84774 

•83835 

•  55968 

.82871 

58 

3 

•50076 

.86559 

•51579 

.85672 

.53066 

•84759 

•54537 

•838*9 

•55992 

•82855 

57 

4 

.50101 

.86544 

.51604 

•85657 

.5309* 

•84743 

•5456i 

.83804 

.56016 

.82839 

56 

5 

.50126 

•86530 

.51628 

.85642 

•53**5 

.84728 

•54586 

.83788 

.56040 

.82822 

55 

6 

•50*5* 

•86515 

•5*653 

.85627 

.847I2 

.54610 

.83772 

.56064 

.82806 

54 

7 

.50176 

.86501 

.51678 

.85612 

•53*64 

.84697 

•54635 

•83756 

.56088 

.82790 

53 

8 

.50201 

.86486 

•  5*7°3 

•85597 

•53*89 

.84681 

•54659 

•83740 

.56112 

.82773 

9 

.50227 

.86471 

.51728 

•85582 

•532*4 

.84666 

•  54683 

•83724 

•56*36 

•82757 

5* 

10 

.50252 

.86457 

•5*753 

•85567 

•53238 

.84650 

.54708 

.83708 

.56160 

.82741 

5° 

ii 

•50277 

.86442 

•5*778 

•8555* 

•53263 

•84635 

•54732 

.83692 

.56184 

.82724 

49 

12 

•  50302 

.86427 

•5*803 

.85536 

•53288 

.84619 

•54756 

•83676 

.  56208 

.82708 

48 

T3 

.86413 

.51828 

•8552* 

•533*2 

.84604 

•5478i 

.83660 

.56232 

.82692 

47 

14 

•50352 

.86398 

.51852 

.85506 

•53337 

.84588 

.54805 

•83645 

.56256 

.82675 

46 

15 

•50377 

.86384 

•5*877 

.85491 

•5336* 

•84573 

.54829 

.83629 

.  56280 

.82659 

45 

16 

•50403 

.86369 

•5*902 

•85476 

•53386 

•84557 

•54854 

.83613 

•56305 

.82643 

44 

17 

.50428 

•86354 

.5*927 

85461 

•534** 

.84542 

•54878 

•83597 

.56329 

.82626 

43 

18 

•50453 

.86340 

.5*952 

.85446 

•53435 

.84526 

.83581 

•56353 

.82610 

42 

*9 

.50478 

.86325 

•5*977 

•8543* 

.53460 

.8451* 

•54927 

•83565 

•56377 

.82593 

20 

•50503 

.86310 

.52002 

.85416 

.53484 

.84495 

•5495* 

•83549 

.56401 

•82577 

40 

21 

.50528 

•862Q5 

.52026 

.85401 

•53509 

.84480 

*•  54975 

•83533 

.56425 

.82561 

39 

22 

•50553 

.86281 

.52051 

•85385 

•53534 

.84464 

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•835*7 

.56449 

•82544 

38 

23 

•50578 

.86266 

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•53558 

.84448 

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.56473 

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37 

24 

.  50603 

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•84433 

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36 

25 

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.84417 

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•82495 

35 

26 

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.52151 

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.  84402 

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.82478 

34 

27 

.50679 

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•52*75 

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•53656 

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.55121 

•83437 

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33 

28 

.50704 

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.82446 

32 

29 

.  50729 

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31 

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.86163 

.85264 

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30 

31 

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.86148 

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29 

32 

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•  52299 

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.82380 

28 

33 

.  50829 

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.82363 

27 

34 

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•5529* 

.83324 

•56736 

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26 

35 

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•83308 

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25 

36 

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24 

37 

.  50929 

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23 

38 

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•  53926 

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22 

39 

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.83244 

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21 

40 

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20 

4i 

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.86000 

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•  5546o 

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42 

•5*°54 

.85985 

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•  54024 

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18 

43 

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44 

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•  54073 

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16 

45 

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IS 

46 

•5**54 

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•83*3* 

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47 

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48 

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12 

49 

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ii 

50 

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10 

5, 

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9 

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8 

53 

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.52819 

.84913 

•54293 

.83978 

•5575° 

•83017 

•57*9* 

.82032 

7 

54 

•5*354 

.85806 

.52844 

.84897 

•543*7 

.83962 

•55775 

.83001 

•572*5 

.82015 

6 

55 

.85792 

.52869 

.84882 

•54342 

.83946 

•55799 

.82985 

•57238 

.81999 

5 

56 

•5*404 

•85777 

•52893 

.84866 

.54366 

.83930 

•55823 

.82969 

.57262 

.81982 

4 

57 

•5*429 

•85762 

•529*8 

.8485* 

•5439* 

•839*5 

.55847 

.82953 

.57286 

.81965 

3 

58 

•5*454 

•85747 

•52943 

.84836 

•544*5 

.83899 

•5587* 

.82936 

•573*o 

.81949 

2 

59 

•5*479 

•85732 

•52967 

.84820 

•  54440 

.83883 

.55895 

.82920 

•57334 

.81932 

I 

60 

•5*504 

•857*7 

•52092 

.84805 

.54464 

.83867 

•559*9 

.82904 

•57358 

.81915 

O 

, 

Cosine 

Sine 

Cosine 

Sine 

Cosine 

Sine 

Cosine 

Sine 

Cosine 

Sine 

, 

59° 

58° 

57° 

56° 

55° 

84 


NATURAL  SINES  AND  COSINES. 


35° 

36° 

37° 

38° 

39° 

Sine 

Cosine 

Sine 

Cosine 

Sine 

Cosine 

Sine 

Cosine 

Sine 

Cosine 

0 

I 

•57358 
•57381 

.81915 
.81899 

•58779 
.58802 

.80902 
.80885 

.60182 
.60205 

.79864 
.79846 

.61566 
•61589 

.78801 
.78783 

.62932 
•62955 

•777*5 
.7/696 

60 

59 

2 

•574°5 

.81882 

.  58826 

.80867 

.60228 

.79829 

.61612 

•78765 

.62977 

•77678 

58 

3 

•57429 

.81865 

•58849 

.  80850 

.60251 

.79811 

•61635 

•78747 

.63000 

.77660 

57 

4 

•57453 

.81848 

•58873 

•80833 

.60274 

•  79793 

.61658 

.78729 

.63022 

.77641 

56 

5 

•57477 

.81832 

.58896 

.80816 

.60298 

.79776 

.61681 

.78711 

•63045 

.77623 

55. 

6 

•57501 

.81815 

.  58920 

.80799 

.60321 

•79758 

.61704 

.78694 

.63068 

•77605 

54 

7 

•57524 

.81798 

•58943 

.80782 

.60344 

•7974* 

.61726 

.78676 

.63090 

.77586 

53 

8 

•57548 

.81782 

•58967 

.80765 

•60367 

•79723 

•61749 

.78658 

•63*13 

.77568 

S2 

9 

•57572 

.81765 

.58990 

.80748 

.60390 

.79706 

.61772 

.78640 

•63*35 

•7755° 

5* 

10 

•57596 

.81748 

.59014 

.80730 

.60414 

.79688 

•6*795 

.  78622 

.63158 

•7753* 

50 

ii 

•57619 

.81731 

•59037 

.80713 

.60437 

.79671 

.61818 

.78604 

.63180 

•775*3 

49 

12 

•57643 

.81714 

.59061 

.80696 

.60460 

•  79653 

.61841 

.78586 

.63203 

•  77494 

48 

13 

.57667 

.81698 

.59084 

.80679 

.60483 

•79635 

.61864 

•78568 

.63225 

•77476 

47 

14 

•  57691 

.81681 

.59108 

.80662 

.  60506 

.79618 

.61887 

•78550 

.63248 

•77458 

46 

15 

•577*5 

.81664 

.5913* 

.80644 

.60529 

.  79600 

.61909 

•78532 

.63271 

•77439 

45 

16 

•57738 

.81647 

•59*54 

.80627 

•60553 

•79583 

.61932 

.78514 

.63293 

.77421 

44 

17 

•57762 

.81631 

•59*78 

.80610 

.60576 

•79565 

•61955 

.78496 

•63316 

.77402 

43 

18 

.57786 

.81614 

.59201 

.80593 

.60599 

•79547 

.61978 

.78478 

•63338 

.77384 

42 

ig 

.57810 

•8*597 

•59225 

.80576 

.60622 

•7953° 

.62001 

.78^60 

.63361 

.77366 

41 

20 

•57833 

.81580 

.59248 

.80558 

.60645 

•795*2 

.  62024 

.78442 

•63383 

•77347 

40 

21 

•57857 

•81563 

•59272 

.80541 

.60668 

.79494 

.  62046 

.78424 

.63406 

•77329 

39 

22 

•57881 

.81546 

•59295 

.80524 

.60691 

•79477 

.  62069 

.78405 

.63428 

•773*0 

38 

23 

•579°4 

•81530 

•593*8 

•80507 

.60714 

•  79459 

.62092 

•78387 

•6345* 

.77292 

37 

24 

.57928 

•8*5*3 

.59342 

.80489 

•60738 

.79441 

.62115 

•78369 

•63473 

•77273 

36 

25 

•57952 

.81496 

•59365 

.80472 

.60761 

.79424 

.62138 

•7835* 

•  63496 

•77255 

35 

26 

•57976 

•8*479 

•59389 

•80455 

.60784 

.79406 

.62160 

•78333 

.635*8 

•77236 

34 

27 

•57999 

.81462 

•594*2 

.80438 

.60807 

.79388 

.62183 

•783*5 

•6354° 

.77218 

33 

28 

.58023 

•8*445 

•59436 

.80420 

.60830 

•7937* 

.62206 

•78297 

•63563 

•77*99 

32 

29 

.58047 

.81428 

•59459 

.80403 

.60853 

•79353 

.62229 

.78279 

•63585 

.77181 

3° 

.58070 

.81412 

.  59482 

.80386 

.60876 

•79335 

.62251 

.78261 

.63608 

.77162 

3° 

31 

•58094 

•8*395 

•595o6 

.80368 

.60899 

•793*8 

.62274 

.78243 

.63630 

•77*44 

29 

S2 

.58118 

.81378 

•59529 

.80351 

.60922 

.79300 

.62297 

.78225 

•63653 

•77*25 

28 

33 

•58141 

.81361 

•59552 

•80334 

.60945 

.79282 

.62320 

.78206 

•63675 

•77*07 

27 

34 

.58165 

•8*344 

•59576 

.80316 

.60968 

.  79264 

•62342 

.78188 

.63698 

.77088 

26 

35 

.58189 

•81327 

•59599 

.80299 

.60991 

.79247 

.62365 

.78170 

.63720 

.77070 

25 

36 

.58212 

.81310 

.59622 

.80282 

.61015 

.79229 

.62388 

.78*52 

•63742 

.77051 

24 

37 

.58236 

•81293 

.  59646 

.  80264 

.61038 

.79211 

.62411 

•78*34 

•63765 

•77°33 

23 

38 

.58260 

.81276 

.59669 

.80247 

.61061 

•79*93 

•62433 

.78116 

•63787 

.77014 

22 

39 

.58283 

.81259 

•59693 

.  80230 

.61084 

•79*76 

.62456 

.78098 

.63810 

.  76996 

21 

40 

•58307 

.81242 

•597*6 

.80212 

.61107 

.79^58 

.62479 

.78079 

.63832 

•  76977 

2O 

4i 

•58330 

.81225 

•59739 

.80195 

.61130 

.79140 

.62502 

.78061 

.63854 

•76959 

*9 

42 

•58354 

.81208 

•59763 

.80178 

•6*153 

.79122 

.62524 

.78043 

.63877 

.76940 

18 

43 

•58378 

.81191 

.59786 

.80160 

.61176 

•79*°5 

.62547 

.78025 

.63899 

.76921 

*7 

44 

.58401 

.81174 

•39809 

.80143 

.61199 

.79087 

.62570 

.78007 

.63922 

•76903 

16 

45 

•58425 

.81157 

•59832 

.80125 

.61222 

.79069 

.62592 

.77988 

•63944 

.76884 

*5 

46 

.58449 

.81140 

•59856 

.80108 

.61245 

•7905* 

.62615 

•77970 

.63966 

.76866 

*4 

47 

.58472 

.81123 

.59879 

.80091 

.61268 

•  79033 

.62638 

•77952 

•63989 

.76847 

13 

48 

.58496 

.81106 

•59902 

.80073 

.61291 

.79016 

.62660 

•77934 

.64011 

.76828 

12 

49 
50 

•58519 
•58543 

.81089 
.81072 

.59926 
•59949 

.80056 
.80038 

.61314 
•6*337 

.78998 
.78980 

.62683 
.62706 

•779*6 
.77897 

•64033 
.64056 

.76810 
.76791 

II 

10 

51 

•58567 

.81055 

•59972 

.80021 

.61360 

.78962 

.62728 

•77879 

.64078 

.76772 

9 

52 

•58590 

.81038 

•  59995 

.80003 

•61383 

•78944 

.62751 

.77861 

.64100 

•  76754 

8 

53 

.58614 

.81021 

.60019 

.79986 

.61406 

.78926 

•62774 

•77843 

.64123 

•76735 

7 

54 

•58637 

.81004 

.60042 

.79968 

.61429 

.78908 

.62796 

.77824 

•64*45 

.76717 

6 

55 

.58661 

.80987 

.60065 

•7995* 

•61451 

.78891 

.62819 

.77806 

.64167 

.76698 

5 

56 

.58684 

.80970 

.60089 

•79934 

.61474 

•78873 

.62842 

.77788 

.64190 

.76679 

4 

57 

•58708 

.80953 

.60112 

•799*6 

•6i497 

•78855 

.62864 

.77769 

.64212 

.76661 

3 

58 
59 

•58731 
•58755 

.80936 
.80919 

•60135 
.60158 

.79899 
.79881 

.61520 
•6*543 

.78837 
.78819 

.62887 
.62909 

•7775* 
•77733 

.64234 
.64256 

.76642 
.76623 

2 

I 

60 

•58779 

.80902 

.60182 

.79864 

.61566 

.78801 

.62932 

•777*5 

.64279 

.76604 

o 

, 

Cosine 

Sine 

Cosine 

Sine 

Cosine 

Sine 

Cosine 

Sine 

Cosine 

Sine 

, 

54° 

53°   1   52° 

51° 

50° 

NATURAL  SINES'  AND  COSINES. 


85 


4 

D° 

4 

[° 

4 

2° 

4, 

5° 

4^ 

0 

n 

Sine 

Cosine 

Sine 

Cosine 

Sine 

Cosine 

Sine 

Cosine 

Sine 

Cosine 

o 

I 

.64279 
•64301 

.76604 
.76586 

.65606 
.65628 

•7547* 
•75452 

.66913 
•66935 

•743*4 
•74295 

.68200 
.68221 

•73*35 
•73*i6 

.69466 
.69487 

.71914 

60 

59 

2 

•64323 

•76567 

.65650 

•75433 

.66956 

•74276 

.68242 

.73096 

.69508 

•7*894 

58 

3 

.64346 

.76548 

.65672 

•754*4 

.66978 

•74256 

.68264 

.73076 

.69529 

•7*873 

57 

4 

.64368 

•7653° 

.65694 

•75395 

.66999 

•74237 

.68285 

•  73056 

•69549 

•7*853 

56 

5 

.64390 

.76511 

•657*6 

•75375 

.67021 

.74217 

.68306 

•73036 

.69570 

•7*833 

55 

6 

.64412 

.76492 

•65738 

•75356 

.67043 

.74198 

.68327 

•730*6 

.69591 

•7*8*3 

54 

7 

•64435 

•76473 

•65759 

•75337 

.67064 

.74178 

.68349 

.  72996 

.69612 

•7*792 

53 

8 

•64457 

•76455 

.65781 

.67086 

•74*59 

.68370 

.72976 

•69633 

•7*772 

52 

9 

•64479 

•76436 

.65803 

•75299 

.67107 

•74*39 

.68391 

•72957 

.69654 

•7*752 

5* 

10 

.64501 

.76417 

.65825 

•75280 

.67129 

.74120 

.68412 

•72937 

.69675 

•7*732 

So 

i 

•64524 

.76398 

.65847 

.75261 

.67151 

.74100 

.68434 

•729*7 

.69696 

.71711 

49 

2 

•64546 

.76380 

.65869 

•7524* 

.67172 

.  74080 

•68455 

•72897 

•697*7 

.71691 

48 

3 

.64568 

.76361 

.65891 

.  75222 

•67*94 

.74061 

.68476 

.72877 

•69737 

.71671 

47 

4 

.64590 

•76342 

•659*3 

.75203 

.67215 

.74041 

.68497 

•72857 

•69758 

•7*650 

46 

.64612 

•76323 

•65935 

•75*84 

•67237 

.74022 

.68518 

•72837 

.69779 

•7*630 

45 

6 

•64635 

.76304 

•65956 

•75*65 

•67258 

.74002 

•68539 

.72817 

.  69800 

.71610 

44 

7 

.64657 

.76286 

•65978 

•75*46 

.67280 

•73983 

.68561 

•72797 

.69821 

•7*590 

43 

8 

.64679 

.76267 

.66000 

•75*26 

.67301 

•  73963 

.68582 

•72777 

.69842 

•7*569 

42 

9 

.64701 

.76248 

.66022 

•75*°7 

•67323 

•73944 

.68603 

•72757 

.69862 

•7*549 

4* 

0 

•64723 

.76229 

.66044 

•75088 

•67344 

•73924 

.68624 

•72737 

.69883 

•7*529 

4° 

T 

.64746 

.76210 

.66066 

.75069 

•67366 

•73904 

.68645 

.72717 

.69904 

•7*508 

39 

2 

3 

.64768 
.64790 

.76192 
•76*73 

.66088 
.66109 

•75050 
•7503° 

•67387 
.67409 

•73885 
•73865 

.68666 
.68688 

•  72697 
•72677 

•69925 
.69946 

.7*488 
.71468 

38 
37 

4 

.64812 

•76*54 

.66131 

.75011 

.67430 

•73846 

.68709 

•72657 

.69966 

•7*447 

36 

.64834 

•76*35 

•66153 

•74992 

•67452 

.73826 

•68730 

•72637 

.69987 

•7*427 

35 

6 

.64856 

.76116 

•66175 

•74973 

•67473 

•73806 

.6875* 

.72617 

.70008 

•7*407 

34 

27 

.64878 

.76097 

.66197 

•74953 

•67495 

•73787 

.68772 

•72597 

.70029 

.71386 

33 

28 

.64901 

.76078 

.66218 

•74934 

•67516 

•73767 

•68793 

•72577 

.70049 

.7*366 

32 

29 

.64923 

•76059 

.66240 

•749*5 

•67538 

•73747 

.68814 

•72557 

.70070 

•7*345 

31 

3° 

.64945 

.76041 

.66262 

.74896 

•67559 

.73728 

.68835 

•72537 

.70091 

•7*325 

30 

31 

.64967 

.76022 

.66284 

.74876 

.67580 

•737o8 

•68857 

•725*7 

.70112 

•7*3°5 

20 

32 

.64989 

.76003 

.66306 

•74857 

.67602 

.73688 

.68878 

.72497 

.70132 

.71284 

28 

33 

.65011 

•75984 

.66327 

.74838 

•67623 

•73669 

.68899 

.72477 

•70*53 

.71264 

27 

34 

•65033 

•75965 

•66349 

.74818 

.67645 

•73649 

.  68920 

•72457 

.70174 

•7*243 

26 

35 

•65055 

•75946 

.66371 

•  74799 

.67666 

•73629 

.68941 

•72437 

•70*95 

.71223 

25 

36 

•65077 

•75927 

•66393 

.74780 

.67688 

•736*0 

.68962 

.•724*7 

.70215 

-.71203 

24 

37 

.65100 

.75908 

.66414 

•7476o 

.67709 

•73590 

.68983 

•72397 

.  70236 

.71182 

23 

38 

.65122 

.75889 

.66436 

•7474* 

•6773° 

•73570 

.69004 

•72377 

.70257 

.71162 

22 

39 

•65*44 

•7587° 

.66458 

.74722 

•67752 

•7355* 

.69025 

•72357 

.70277 

.71141 

21 

40 

.65166 

•7585* 

.66480 

•74703 

•67773 

•7353* 

.69046 

•72337 

.70298 

.71121 

2O 

4* 

.65188 

•75832 

.66501 

.74683 

•67795 

•735** 

.69067 

•723*7 

•703*9 

.71100 

19 

42 

.65210 

•758*3 

.66523 

.  74664 

.67816 

•7349* 

.69088 

.72297 

•70339 

.71080 

18 

43 

.65232 

•75794 

•66545 

•74644 

•67837 

•73472 

.69109 

.72277 

.70360 

•7*059 

*7 

44 

•65254 

•75775 

.66566 

•74625 

.67859 

•73452 

•69130 

•72257 

.70381 

•7*039 

16 

45 

•65276 

•75756 

.66588 

.74606 

.67880 

•73432 

.69151 

.72236 

.70401 

.71019 

15 

46 

.65298 

•75738' 

.66610 

•74586 

.67901 

•734*3 

.69172 

.72216 

.70422 

.70998 

*4 

47 

.65320 

•757*9 

.66632 

•74567 

•67923 

•73393 

.69193 

.72196 

•7°443 

.70978 

*3 

48 

•65342 

•757°° 

•66653 

.74548 

•67944 

•73373 

.69214 

.72176 

•  70463 

•70957 

12 

49 

•65364 

.75680 

.66675 

.74528 

.67965 

•73353 

•69235 

•72*56 

.70484 

•70937 

II 

50 

.65386 

.75661 

.66697 

•74509 

.67987 

•73333 

•69256 

.72136 

•70505 

.70916 

10 

51 

.65408 

•75642 

.66718 

.74489 

.68008 

•733*4 

.69277 

.72116 

.70525 

.70896 

9 

52 

•6543° 

•75623 

.66740 

.74470 

.68029 

•73294 

.69298 

•72095 

.70546 

•7-875 

8 

53 

•65452 

.75604 

.66762 

•7445* 

.68051 

•73274 

.69319 

•72075 

•  70567 

•70855 

7 

54 

•65474 

•75585 

.66783 

•7443* 

.68072 

•73254 

.69340 

•72055 

.70587 

.70834 

6 

55 

•65496 

•75566 

.66805 

.74412 

.68093 

•73234 

.69361 

•72035 

.70608 

.70813 

5 

56 

.655*8 

•75547 

.66827 

•74392 

.68115 

•732*5 

.69382 

.72015 

.70628 

•70793 

4 

57 

•6554° 

•75528 

.66848 

•74373 

.68136 

•73*95 

.69403 

•7*995 

.  70649 

•70772 

3 

58 

.65562 

•75509 

.66870 

•74353 

•68157 

•73*75 

.69424 

•7*974 

.70670 

.70752 

2 

59 

•65584 

•75490 

.66891 

•74334 

.68179 

•73*55 

•69445 

•7*954 

.70690 

•7°73* 

I 

60 

.65606 

•7547* 

.66913 

•743*4 

.68200 

•73*35 

.69466 

•7*934 

.70711 

.70711 

O 

, 

Cosine 

Sine 

Cosine 

Sine 

Cosine 

Sine 

Cosine 

Sine 

Cosine 

Sine 

-, 

4 

?° 

4* 

4 

7° 

4 

5° 

4. 

.0 

3 

86 


NATURAL  TANGENTS  AND  COTANGENTS. 


< 

>° 

] 

i 

5° 

: 

,° 

i 

^° 

Tang 

Cotang 

Tang 

Cotang 

Tang 

Cotang 

i 

Tang 

Cotang 

Tang 

Cotang 

o 

.00000 

Infin. 

.01746 

57.2900 

.03402 

28.6363 

.05241 

19.0811 

.06993 

14.3007 

60 

I 

.00029 

3437-75 

•01775 

56.3506 

.03521 

28.3994 

•05270 

i8.9755 

.07022 

14.2411 

59 

2 

.00058 

1718.87 

.01804 

55-4415 

-0355° 

28.1664 

•05299 

18.8711 

.07051 

14.  1821 

58 

3 

.00087 

1145.92 

•01833 

•03579 

27.9372 

.05328 

18.7678 

.07080 

14-1235 

57 

4 

.00116 

859-436 

.01862 

53-7086 

.03609 

27.7117 

•05357 

18.6656 

.07110 

14.0655 

56 

5 

.00145 

687.549 

.01891 

52.8821 

-03638 

27.4899 

•05387 

18.5645 

•07139 

14.0079 

55 

6 

.00175 

572-957 

.01920 

52.0807 

-03667 

27.2715 

.05416 

18.4645 

.07168 

13-9507 

54 

7 

.00204 

491.106 

.01949 

51.3032 

.03696 

27.0566 

•05445 

18-3655 

.07197 

13.8940 

53 

8 

.00233 

429.718 

.01978 

50-5485 

.03725 

26.8450 

•05474 

18.2677 

.07227 

13-8378 

52 

9 

.00262 

381.971 

.02007 

49.8157 

•03754 

26.6367 

.05503 

18.1708 

.07256 

13.7821 

Si 

10 

.00291 

343-774 

.  02036 

49.1039 

•03783 

26.4316 

•05533 

18.0750 

.07285 

13.7267 

50 

ii 

.00320 

312.521 

.02066 

48.4121 

.03812 

26.2296 

.05562 

17.9802 

•07314 

13.6719 

49 

12 

.00349 

286.478 

.02095 

47-7395 

.03842 

26.0307 

•0559i 

17.8863 

•07344 

13.6174 

48 

13 

.00378 

264.441 

.02124 

47.08,53 

•03871 

25.8348 

.05620 

17-7934 

•°7373 

13-5634 

47 

15 

.00407 
.00436 

245-552 
229.  182 

•02153 
.02182 

46-4489 
45.8294 

.03900 
.03929 

25.6418 
25-4517 

.05649 
.05678 

17.7015 
17.6106 

.07402 
.07431 

13.5098 
13-4566 

46 
45 

16 

.00465 

214.858 

.O22II 

45.2261 

.03958 

25.2644 

.05708 

17-5205 

.07461 

13-4039 

44 

17 

.00495 

202.219 

.O224O 

44.6386 

.03987 

25-0798 

•05737 

17-4314 

.07490 

i3-35I5 

43 

18 

.00524 

190.984 

.02269 

44.0661 

.04016 

24.8978 

.05766 

17-3432 

•07519 

13.2996 

42 

19 

•oo553 

180.932 

.02298 

43.5081 

.  04046 

24.7185 

•05795 

17-2558 

.07548 

13.2480 

20 

.00582 

171.885 

.02328 

42.9641 

.04075 

24.5418 

.05824 

17.1693 

.07578 

13-1969 

40 

21 

.00611 

163.700 

•02357 

42.4335 

.04104 

24-3675 

•05854 

17.0837 

.07607 

13.1461 

39 

22 

.00640 

156-259 

.02386 

41.9158 

•04133 

24.1957 

•05883 

16.9990 

.07636 

13.0958 

38 

23 

.00669 

149-465 

.02415 

41.4106 

.04162 

24.0263 

.05912 

16.9150 

.07665 

13.0458 

37 

24 

.00698 

143-237 

.02444 

40.9174 

.04191 

23-8593 

.05941 

16.8319 

•07695 

12.9962 

36 

25 

.00727 

137-507 

•02473 

40.4358 

.04220 

23.6945 

•0597° 

16.7496 

•07724 

12.9469 

35 

26 

•00756 

132.219 

.O25O2 

39-9655 

.04250 

23-5321 

•05999 

16.6681 

•07753 

12.8981 

34 

27 

.00785 

127.321 

.02531 

39.5059 

•04279 

.06029 

16.5874 

.07782 

12.8496 

33 

28 

.00815 

122.774 

.02560 

39.0568 

.04308 

23.2137 

.06058 

.07812 

12.8014 

S2 

29 

.00844 

118.540 

.02589 

38.6177 

•04337 

23-0577 

.06087 

16.4283 

.07841 

I2-7536 

30 

•00873 

114.589 

.02619 

38.1885 

.04366 

22  .9038 

.06116 

16-3499 

.07870 

I  2  .  7062 

3° 

31 

.00902 

110.892 

.02648 

37-7686 

•04395 

22.7519 

.06145 

16.2722 

.07899 

12  .6591 

29 

32 

.00931 

107.426 

.02677 

37-3579 

.04424 

22.6O2O 

.06175 

16.1952 

.07929 

12.6l24 

28 

33 

.00960 

104.171 

.02706 

36-9560 

•04454 

22.4541 

.06204 

16.1190 

.07958 

12.5660 

27 

34 

.00989 

101.  107 

•02735 

36.5627 

.04483 

22.3081 

.06233 

16.0435 

.07987 

12.5199 

26 

35 

.01018 

98.2179 

.02764 

36.1776 

•04512 

22.  1640 

.06262 

15-9687 

.08017 

12.4742 

25 

36 

.01047' 

95.4895 

.02793 

35-8006 

.04541 

2     .0217 

.06291 

15-8945 

.08046 

12.4288 

24 

37 

.01076 

92.9085 

.O2822 

35.4313 

.04570 

2     .8813 

.06321 

15.8211 

.08075 

12.3838 

23 

38 

.01105 

90.4633 

.02851 

35-o695 

.04599 

2     .7426 

-06350 

15-7483 

.08104 

12.3300 

22 

39 

•01135 

88.1436 

.02881 

34.7151 

.04628 

2     .6056 

-06379 

15.6762 

.08134 

12.2946 

21 

40 

.01164 

85.9398 

.02910 

34-3678 

.04658 

2     .4704 

.06408 

15.6048 

.08163 

12.2505 

2O 

4i 

.01193 

83.8435 

.02939 

34.0273 

.04687 

21.3369 

•06437 

15-5340 

.08192 

12.2067 

J9 

42 

.01222 

81.8470 

.02968 

33.6935 

.04716 

21.2049 

.06467 

15.4638 

.08221 

12.1632 

18 

43 

.01251 

79-9434 

.02997 

33.3662 

•04745 

21.0747 

.06496 

15-3943 

.08251 

J2.I20I 

17 

44 

.01280 

78.  1263 

.  03026 

33-0452 

•04774 

20.9460 

.06525 

15-3254 

.08280 

12.0772 

16 

45 

.01309 

76.3900 

•03055 

32.7303 

.  04803 

20.8l88 

-06554 

15-2571 

.08309 

12.0346 

15 

46 

.01338 

74.7292 

.03084 

32.4213 

.04833 

20.6932 

.06584 

15-1893 

•08339 

11.9923 

M 

47 

.01367 

73.!39o 

.03114 

32.1181 

.04862 

20.5691 

.06613 

15.1222 

.08368 

11.9504 

48 

.01396 

71.6151 

.03143 

31.8205 

.04891 

20.4465 

.06642 

15-0557 

.08397 

II.9087 

12 

49 

.01425 

70.1533 

.03172 

31-5284 

.04920 

20.3253 

.06671 

14.9898 

.08427 

11.8673 

II 

50 

-01455 

68.7501 

.03201 

31.2416 

.04949 

2O.2O56 

.06700 

14-9244 

.08456 

11.8262 

IO 

5i 

.01484 

67.4019 

.03230 

30.9599 

.04978 

20.0872 

.06730 

14-8596 

08485 

I1.7853 

9 

S2 

•01513 

66.  1055 

.03259 

30-6833 

•05007 

19.9702 

.06759 

14-7954 

.08514 

11.7448 

8 

53 

.01542 

64.8580 

.03288 

30.4116 

•05037 

19-8546 

.06788 

14-7317 

.08544 

11.7045 

7 

54 

.01571 

63-6567 

.03317 

30.1446 

.05066 

19.7403 

.06817 

14.6685 

-08573 

11.6645 

6 

55 

.01600 

62.4992 

.03346 

29.8823 

•05095 

19.6273 

.06847 

14.6059 

.08602 

11.6248 

5 

56 

.01629 

61.3829 

•03376 

29.6245 

.05124 

19.5156 

.06876 

14-5438 

.08632 

H.5853 

4 

57 

.01658 

60.3058 

•03405 

29-3711 

•05153 

19.4051 

.06905 

14.4823 

.08661 

11.5461 

3 

58 

.01687 

59-2659 

•03434 

29.1220 

.05182 

19.2959 

.06034 

14.4212 

.08690 

11.5072 

2 

59 

.01716 

58.2612 

•03463 

28.8771 

.05212 

19.1879 

.06963 

14.3607 

.08720 

11.4685 

I 

60 

.01746 

57.2900 

.034Q2 

28.6363 

-05241 

I9.o8ll 

.06993 

14.3007 

.08749 

11.4301 

O 

Cotang 

Tang 

Cotang 

Tang 

Cotang 

Tang 

Cotang 

Tang 

Cotang 

Tang 

' 

1 

8< 

)° 

8* 

B 

8; 

8^ 

)° 

8c 

,o 

NATURAL  TANGENTS  AND  COTANGENTS. 


87 


5 

0 

6 

0 

7°       1 

g 

o 

9 

0 

Tang 

Cotang 

Tang 

Cotang 

Tang 

Cotang 

Tang 

Cotang 

Tang 

Cotang 

o 

.08749 

11.4301 

.  10510 

9-  51430 

.12278 

8.14435 

.14054 

7-11537 

-15838 

6-31375 

60 

I 

.08778 

11.3919 

.  10540 

9.48781 

.12308 

8.12481 

.14084 

7.  10038 

.15868 

6.30189 

59 

2 

'  .08807 

"•3540 

.  10569 

0.46141 

.  12338 

8.  10536 

.14113 

7.08546 

.15898 

6.29007 

58 

3 

.08837 

11-3163 

•  10599 

9-435I5 

.12367 

8.08600 

•14143 

7.07059 

-15928 

6.27829 

57 

4 

.08866 

11.2789 

.  10628 

9.40904 

.12397 

8.06674 

.14173 

7-05579 

•15958 

6.26655 

56 

5 

.08895 

11.2417 

.10657 

9  •  38307 

.12426 

8.04756 

.  14202 

7.04105 

•15988 

6.25486 

55 

6 

.08925 

ii  .2048 

.10687 

9-35724 

.12456 

8.02848 

.14232 

7.02637 

.  16017 

6.24321 

54 

7 

.08954 

n.  1681 

.10716 

9-33155 

.12485 

8.00948 

.  14262 

7.01174 

.16047 

6.23160 

53 

8 

.08983 

ii  .  1316 

.  10746 

9-30599 

•12515 

7.99058 

.14291 

6.99718 

.16077 

6.22003 

S2 

9 

.09013 

11.0954 

•I0775 

9.28058 

•12544 

7-97176 

.14321 

6.98268 

.  16107 

6  .  2085  i 

51 

10 

.09042 

11.0594 

.10805 

9-2553° 

.12574 

7-95302 

•14351 

6.96823 

.16137 

6.19703 

So 

ii 

.09071 

11.0237 

.10834 

9.23016 

.12603 

7-93438 

.14381 

6.95385 

.  16167 

6.18559 

40 

12 

.09101 

10.9882 

.10863 

9.20516 

,  12633 

7-91582 

.14410 

6.93952 

.16196 

6.17419 

48 

13 

.09130 

10.9529 

.  10893 

9.18028 

.12662 

7-89734 

.  14440 

6.92525 

.16226 

6.16283 

47 

14 

.09159 

10.9178 

.  10922 

9-15554 

.  12692 

7.87895 

.14470 

6.91104 

.16256 

6.15151 

46 

15 

.09189 

10.8829 

.10952 

9.13093 

.12722 

7.86064 

•14499 

6  .  89688 

.16286 

6.14023 

45 

16 

.09218 

10.8483 

.10981 

9  .  10646 

.12751 

7.84242 

•14529 

6.88278 

.16316 

6.12899 

44 

17 

.09247 

10.8139 

.  J'lOII 

9.08211 

.  12781 

7.82428 

•14559 

6.86874 

•  16346 

6.11779 

43 

18 

.09277 

10.7797* 

.  1  1040 

9.05789 

.12810 

7.80622 

.14588 

6.85475 

•16376 

6.10664 

42 

19 

.09306 

10.7457 

.  i  1070 

9-03379 

.  12840 

7.78825 

.14618 

6.84082 

.16405 

6.09552 

4i 

20 

•09335 

10.7119 

.11099 

9.00983 

.  12869 

7-77035 

.14648 

6.82694 

^6435 

6.08444 

40 

21 

•  °9365 

10.6783 

.11128 

8.98598 

.12899 

7-75254 

.14678 

6.81812 

.  16465 

6.07340 

39 

22 

.09394 

10.6450 

.11158 

8.96227 

.12929 

7.73480 

.14707 

6  .  79936 

.16495 

6.06240 

38 

23 

.09423 

10.6118 

.11187 

8.93867 

.12958 

7-  7*715 

•14737 

6.78564 

.16525 

6.05143 

37 

24 

•09453 

10.5789 

.11217 

8.91520 

.12988 

7.69957 

.14767 

6.77199 

•16555 

6.04051 

36 

25 

.09482 

10.5462 

.11246 

8.89185 

.13017 

7.68208 

•14796 

6-758JB 

-16585 

6.02962 

35 

26 

.09511 

10.5136 

.  11276 

8.86862 

•13047 

7.66466 

.14826 

6.74483 

.16615 

6.01878 

34 

27 

.09541 

10.4813 

.11305 

8.84551 

.13076 

7.64732 

-14856 

6-  73  '33 

.16645 

6.00797 

33 

28 

.09570 

10.4491 

•"335 

8.82252 

.13106 

7.63005 

.14886 

6.71789 

.16674 

5-99720 

S2 

29 

.09600 

10.4172 

.11364 

8.79964 

•13136 

7.61287 

•149*5 

6  .  7045° 

.16704 

5-98646 

31 

3° 

.09629 

10.3854 

.11394 

8.77689 

•13165 

7-59575 

.14945 

6.69116 

•16734 

5-97576 

30 

31 

.09658 

10.3538 

.11423 

8.75425 

•i3J95 

7.57872 

•14975 

6.67787 

.  16764 

5-9651° 

29 

3  2 

.09688 

10.3224 

.11452 

8  73172 

.13224 

7.56176 

.15005 

6.66463 

.16794 

5-95448 

28 

33 

.09717 

10.2913 

.11482 

8.70931 

•i3254 

7-54487 

•  15034 

6.65144 

.16824 

5-9439° 

27 

1  34 

.09746 

10.2602 

•  11511 

8.68701 

.13284 

7.52806 

•15064 

6.63831 

.16854 

5-93335 

26 

35 

.09776 

10.2294 

.11541 

8.66482 

•I33I3 

7-5"32 

.15094 

6.62523 

.16884 

5.92283 

25 

36 

.09805 

10.1988 

.11570 

8.64275 

•I3343 

7  •  49465 

.15124 

6.61219 

.  16914 

5.91236 

24 

37 

.09834 

10.1683 

.  11600 

8.62078 

•J3372 

7.47806 

•15153 

6.59921 

•1^944 

5  90191 

23 

I  38 

.09864 

10.1381 

.  11629 

8.59893 

•  13402 

7-46i54 

•15183 

6.58627 

.16974 

5.89151 

22 

39 

.09893 

10.  1080 

•11659 

8.57718 

•13432 

7-44509 

.15213 

6-57339 

.17004 

5.88114 

21 

40 

.09923 

10.0780 

.11688 

8-55555 

.13461 

7.42871 

•15243 

6.56055 

•I7°33 

5.87080 

2O 

41 

.09952 

10.0483 

.11718 

8.53402 

•13491 

7.41240 

.15272 

6-54777 

.17063 

5.86051 

19 

42 

.09981 

10.0187 

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8.51259 

•I.352I 

7.39616 

•15302 

6.535°3 

•17093 

5.85024 

18 

43 

.    OOII 

9.98931 

.11777 

8.49128 

•1355° 

7-37999 

•15332 

6.52234 

.17123 

5.84001 

J7 

44 

.  0040 

9.96007 

.  1806 

8.47007 

.13580 

7.36389 

•15362 

6.50970 

•I7I53 

5.82982 

16 

45 

.  0069 

9.93101 

.  1836 

8.44896 

.13609 

7.34786 

•15391 

6.49710 

•17183 

5.81966 

15 

46 

.  0099 

9.90211 

.  1865 

8.42795 

•13639 

7-33I90 

.15421 

6.48456 

.17213 

5-80953 

14 

47 

.  0128 

9-87338 

•  1895 

8.40705 

.13669 

7.31600 

•15451 

6.47206 

•17243 

5-79944 

13 

48 

•   0158 

9.84482 

.  1924 

8.38625 

.  13698 

7.30018 

•15481 

6.45961 

•17273 

5-78938 

12 

49 

.   0187 

9.81641 

.11954 

8.36555 

.13728 

7.28442 

•  155" 

6.44720 

•17303 

5-77936 

ii 

5° 

.  0216 

9.78817 

.  1983 

8.34496 

•'3758 

7.26873 

•1554° 

6.43484 

•17333 

5-76937 

10 

5i 

.  0246 

9.76009 

•   2013 

8.32446 

.13787 

7.25310 

•15570 

6.42253 

.17363 

5-75941 

9 

52 

•  0275 

9.73217 

.    2042 

8  .  30406 

.13817 

7-23754 

.15600 

6.41026 

•17393 

5  •  74949 

8 

53 

•   °3°5 

9.70441 

.  2072 

8.28376 

.13846 

7.22204 

•1563° 

6.39804 

•17423 

5.7396o 

7 

54 

•   0334 

9.67680 

.     2IOI 

8-26355 

.13876 

7.20661 

.15660 

6.38587 

•17453 

5-72974 

6 

55 

.   0363 

9-64935 

.   2131 

8.24345 

.13906 

7.19125 

.15689 

6-37374 

•17483 

5-71992 

5 

56 

•   °393 

9.62205 

.   2160 

8.22344 

•13935 

7-17594 

•15719 

6.36165 

•17513 

5-7IOI3 

4 

57 

.   0422 

9.59490 

.   2190 

8.20352 

.13965 

7.16071 

•15749 

6.34961 

•17543 

5-70°37 

3 

58 

•   0452 

9.56791 

.   2219 

8.18370 

•'3995 

7-14553 

•15779 

6.33761 

•17573 

5.69064 

2 

59 

.  0481 

9.54106 

•   2249 

8.16398 

.14024 

7.13042 

.15809 

6.32566 

.17603 

5-68094 

I 

60 

•   0510 

9.51436 

.   2278 

8.14435 

.14054 

7-II537 

•15838 

6.31375 

•17633 

5.67128 

O 

/ 

Cotang 

Tang 

Cotang 

Tang 

Cotang 

Tang 

Cotang 

Tang 

Cotang 

Tang 

/ 

8 

4° 

8 

3° 

8 

2° 

8 

1° 

8 

D° 

88 


NATURAL  TANGENTS  AND  COTANGENTS. 


I 

0° 

I 

i° 

I 

2 

I 

3° 

i 

4° 

Tang 

Cotang 

Tang 

Cotang 

Tang 

Cotang 

Tang 

Cotang 

Tang 

Cotang 

o 

•i7633 

5.67128 

•  19438 

5-14455 

.21256 

4.70463 

•23087 

4-33148 

-24933 

4.01078 

60 

I 

.17663 

5.66165 

.19468 

5-i365« 

.21286 

.69791 

.23117 

4-32573 

.24064 

4.00582 

59 

2 

•17693 

5-65205 

.  19498 

5.12862 

.21316 

.23148 

.32001 

•  24995 

4.00086 

58 

3 

•17723 

5.64248 

•19529 

5.12069 

•21347 

'.  1  68452 

•23179 

•3H30 

.25026 

3-99592 

57 

4 

•17753 

5-63295 

•19559 

5.11279 

•21377 

.67786 

•23209 

.  30860 

.25056 

3.99099 

56 

5 

•17783 

5-62344 

.19589 

5.10490 

.21408 

.67121 

.23240 

.30291 

.2=5087 

3.98607 

55 

6 

.17813 

5.61307 

.19619 

5.09704 

.21438 

.66458 

.23271 

.29724 

.25118 

3.98117 

54 

7 

•17843 

5.60452 

.  19649 

5.08921 

.21469 

•65797 

.23301 

•29*59 

•25149 

3.97627 

53 

8 
9 

•17873 
•17903 

5-595" 
5-58573 

.  19680 
.19710 

5.08139 
5.07360 

.21499 
•21529 

.65138 
.64480 

•23332 
•23363 

•28595 
.28032 

.25180 
•25211 

3-97139 
3.96651 

52 
51 

10 

•J7933 

5-57638 

•  1974° 

5.06584 

.21560 

4.63825 

•23393 

.27471 

.25242 

3.96165 

So 

ii 

•i7963 

5.56706 

.19770 

5.05809 

.21590 

4.63171 

.23424 

.26911 

•  252.73 

3.95680 

49 

12 

•17993 

5-55777 

.  19801 

5-05037 

.21621 

4.62518 

•23455 

.26352 

-25304 

3.95196 

48 

13 

.18023 

5-54851 

.19831 

5.04267 

.21651 

4.61868 

•23485 

•25795 

-25335 

3-94713 

47 

14 

•18053 

5-53927 

.19861 

5-03499 

.21682 

4.61219 

.23516 

•25239 

.25366 

3-94232 

46 

15 

.  18083 

5-53°°7 

.19891 

5-02734 

.21712 

4.60572 

•23547 

.24685 

•25397 

3-93751 

45 

16 

.18113 

5-52090 

.19921 

5.01971 

•21743 

4-59927 

-23578 

.24132 

.25428 

3-93271 

44 

17 

.18143 

S-Sii?6 

•  19952 

5.01210 

•21773 

4-59283 

.23608 

.23580 

•25459 

3-92793 

43 

18 

.18173 

5.50264 

.19982 

5.00451 

.  2  I  804 

4.58641 

•  23639 

•23030 

•  25490 

3.92316 

42 

19 

.18203 

5-49356 

.20012 

4-99^95 

.21834 

4.58001 

.23670 

.22481 

•25521 

3-91839 

41 

20 

•18233 

5-4845I 

.2OO42 

4.98940 

.21864 

4-57363 

.23700 

4-21933 

•25552 

3.91364 

40 

21 

.  18263 

5-47548 

.20073 

4.98188 

.21895 

4.56726 

•23731 

4.21387 

•25583 

3-90890 

39 

22 

•  I&293 

5-46648 

.2OIO3 

4-97438 

.21925 

4.56091 

•25762 

4.20842 

.25614 

3.90417 

38 

23 

.18323 

5-45751 

.20133 

4.96690 

.21956 

4-55458 

•23793 

4.20298 

•25645 

3-89945 

37 

24 

•18353 

5-44857 

.20164 

4-95945 

.21986 

4-54826 

•23823 

4.19756 

.25676 

3-89474 

36 

25 

•  18384 

5-43966 

1  90194 

4.95201 

.22017 

4.54196 

•23854 

4.19215 

.25707 

3  .  89004 

35 

26 

.18414 

5-43077 

.20224 

4.94460 

.22047 

4-53568 

•23885 

4.18675 

-25738 

3-88536 

34 

27 

.18444 

5.42192 

.20254 

4-93721 

.22078 

4.52941 

.23916 

4.18137 

.2,5769 

3.88068 

33 

28 

.18474 

5.41309 

.20285 

4.92984 

.22108 

4.52316 

.23946 

4.17600 

.25800 

3.87601 

32 

29 

.18504 

5.40429 

•20315 

4.92249 

•22139 

4.51693 

•23977 

4.17064 

.25831 

3-87136 

31 

30   1 

.18534 

5-39552 

•  20345 

4.91516 

.22169 

4.51071 

.24008 

4.16530 

.25862 

3.86671 

30 

31 

.18564 

5.38677 

•20376 

4.90785 

.222OO 

4.50451 

•24039 

4-15997 

.25893 

3  .  86208 

29 

32 

•  18594 

5-378o5 

.  20406 

4.90056 

.22231 

4-49832 

.24069 

4.15465 

-25924 

3-85745 

28 

33 

.18624 

5-36936 

.20436 

4.89330 

.2226l 

4.49215 

.24100 

4-14934 

•25955 

3.85284 

27 

34 

.18654 

5.36070 

.20466 

4.88605 

.22292 

4.48600 

.24131 

4.14405 

.25986 

3.84824 

26 

35 

.18684 

5-35206 

.20497 

4.87882 

.22322 

4-47986 

.24162 

4-13877 

.26017 

3-84364 

25 

36 

.18714 

5-34345 

.20527 

4.87162 

•22353 

4-47374 

•24193 

4-I3350 

.26048 

3.83906 

24 

37 

•18745 

5-33487 

•20557 

4.86444 

•22383 

4.46764 

•24223 

4.12825 

.26079 

3-83449 

23 

138 

•18775 

5-32631 

.20588 

4-85727 

.22414 

4-46i55 

•24254 

4.12301 

.26110 

3.82992 

22 

39 

.18805 

5-3r778 

.20618 

4.85013 

.22444 

4-45548 

-24285 

4.11778 

.26141 

3-82537 

21 

40 

•18835 

5.30928 

.20648 

4.84300 

•22475 

4.44942 

.24316 

4-11256 

.26172 

3.82083 

2O 

41 

.18865 

5.30080 

.20679 

4-83590 

•  22505 

4-44338 

•24347 

4.10736 

.  26203 

3.81630 

iq 

42 

.18895 

5-29235 

.20709 

4.82882 

.22536 

4-43735 

•24377 

4.10216 

•26235 

3-8ii77 

18 

43 

•18925 

5-28393 

.20739 

4-82175 

•22567 

4-43T34 

.24408 

4.09699 

.26266 

3.80726 

17 

44 

•18955 

5-27553 

.20770 

4.81471 

.22597 

4-42534 

•24439 

4.09182 

.26297 

3.80276 

16 

45 

.18986 

5-26715 

.20800 

4.80769 

.22628 

4.41936 

.24470 

4.08666 

.26328 

3.79827 

15 

46 

.  IgOl6 

5.25880 

.20830 

4.80068 

.22658 

4.41340 

•24501 

4.08152 

.26359 

3-79378 

14 

47 

.  19046 

5-25048 

.20861 

4-7937° 

.22689 

4-40745 

•24532 

4-07639 

.26390 

3-78931 

i.3 

48 

.  19076 

5.24218 

.20891 

4.78673 

.22719 

4.40152 

.24562 

4.07127 

.26421 

3-78485 

12 

49 

.  19106 

5-23391 

.20921 

4.77978 

.22750 

4-3956o 

•24593 

4.06616 

.26452 

3.78040 

11 

5° 

.I9I3"6 

5-22566 

.  20952 

4.77286 

.22781 

4.38969 

.24624 

4.06107 

.26483 

3-77595 

IO 

5i 

.19166 

5-21744 

.20982 

4.76595 

.22811 

4-38381 

•24655 

4-05599 

-26515 

3-77I52 

9 

52 

.19197 

5-20925 

•2    013 

4-759o6 

.22842 

4-37793 

.24686 

4.05092 

.26546 

3.76709 

8 

53 

.19227 

5.20107 

.2    043 

4-752I9 

.22872 

4.37207 

.24717 

4.04586 

•26577 

3.76268 

7 

54 

•19257 

5-I9293 

.2    073 

4-74534 

.22903 

4.36623 

.24747 

4  .  0408  i 

.26608 

3.75828 

6 

55 

.19287 

5.18480 

.2    104 

4-7385I 

•22934 

4.36040 

.24778 

4-03578 

.26639 

3.75388 

5 

56 

•I93I7 

5.17671 

.2    I34 

4-73I70 

.22964 

4-35459 

.  24809 

4-03076 

.  26670 

3-  7495° 

4 

57 

•19347 

5.16863 

.2    l64 

4.72490 

.22995 

4-34879 

.24840 

4.02574 

.26701 

3-74512 

3 

58 

•19378 

5-16058 

•2    195 

4.71813 

.23026 

4-34300 

.24871 

4.02074 

•26733 

3  •  74075 

2 

59 

.19408 

5-  15256 

.21225 

4-7II37 

.  23056 

4-33723 

.24902 

4.01576 

.26764 

3  •  7364° 

I 

60 

.19438 

5-14455 

.21256 

4.70463 

.23087 

4.33148 

•24933 

4.01078 

•26795 

3-73205 

o 

, 

Cotang 

Tang 

Cotang 

Tang 

Cotang 

Tang 

Cotang 

Tang 

Cotang 

Tang 

~| 
l 

7 

9° 

7 

s.° 

7 

1° 

7 

6° 

7 

5° 

NATURAL  TANGENTS  AND  COTANGENTS. 


89 


15° 

16° 

17° 

18° 

'9° 

Tang 

Cotang 

Tang 

Cotang 

Tang 

Cotang 

Tang 

Cotang 

Tang 

Cotang 

0 

.20795 

3-73205 

.28075 

3.48741 

•3°573 

3.27085 

•  32492 

3.07768 

•  34433 

2  .  9042  I 

60 

I 

.26826 

3.72771 

.28706 

3-48359 

.30605 

3-26745 

•32524 

3.07464 

•34465 

2.90147 

59 

2 

.26857 

3-72338 

.28738 

3-47977 

•30637 

3.26406 

32556 

3.07160 

.34498 

2.89873 

58 

3 

.26888 

3.71907 

.28769 

3-47596 

.30669 

3.26067 

.32588 

3.06057 

•3453° 

2.89600 

57 

4 

.26920 

3-7M76 

.28800 

3.47216 

.30700 

3.25729 

.32621 

3  -c6554 

•  34563 

2.89327 

56 

5 

.26951 

3.71046 

.28832 

3-46837 

•3°732 

3-25392 

•32653 

3.06252 

•34596 

2.89055 

55 

6 

.26982 

3.70616 

.28864' 

3-46458 

.30764 

3-25055 

.32685 

3-0595o 

.34628 

2.88783 

54 

7 

.27013 

3.70188 

.28895 

3.46080 

.30796 

3.24719 

•32717 

3-05649 

.34661 

2.88511 

53 

8 

.  27044 

3.69761 

.28927 

3.45703 

.30828 

3  •  24383 

•32749 

3-05349 

•34693 

2  .  88240 

S2 

9 

.27076 

3-69335 

•28958 

3-45327 

.30860 

3.24049 

.32782 

3-05049 

.34726 

2.87970 

51 

10 

.27107 

3.68909 

.28990 

3-  4495  i 

.30891 

3-237I4 

.32814 

3-04749 

•34758 

2.87700 

5° 

n 

.27138 

3.68485 

.29021 

3.44576 

.30923 

3-23381 

.32846 

3.04450 

•34791 

2-87430 

49 

12 

.27169 

3.68061 

•29053 

3.44202 

•3°955 

3.23048 

.32878 

3-o4!52 

•34824 

2.87161 

48 

J3 

.27201 

3.67638 

.29084 

3.43829 

.30987 

3.22715 

.32911 

3-03854 

.34856 

2.86892 

47 

J4 

•27232 

3.67217 

.29116 

3-43456 

.31019 

3.22384 

•32943 

3-03556 

.34889 

2.86624 

46 

15 

.27263 

3.66796 

.29147 

3-43084 

•3*°5l 

3-22053 

•32975 

3.03260 

•34922 

2.86356 

45 

16 

.27294 

3.66376 

.29179 

3-427*3 

.31083 

3.21722 

•33007 

3.02963 

•34954 

2.86089 

44 

J7 

.27326 

3-65957 

.29210 

3-42343 

•Sins 

3.21392 

•33040 

3.02667 

.34987 

2.85822 

43 

18 

•27357 

3-65538 

.29242 

3-4I973 

•3"47 

3.21063 

•33072 

3-02372 

.35020 

2-85555 

42 

'9 

.27388 

3.65121 

•29274 

3.41604 

.31178 

3-20734 

•33I04 

3.02077 

•35052 

2.85289 

41 

20 

.27419 

3-64705 

.29305 

3.41236 

.31210 

3  .  20406 

-33136 

3-01783 

•35085 

2.85023 

40 

21 

•27451 

3.64289 

•29337 

3.40869 

.31242 

3.20079 

•33169 

3.01480, 

.35118 

2.84758 

39 

22 

.27482 

3-63874 

.29368 

3.40502 

•31274 

3-J9752 

•33201 

3.01196 

•35150 

2  .  84494 

38 

23 

•275J3 

3.63461 

.29400 

3.40136 

.31306 

3.19426 

•33233 

3.00903 

•35i83 

2.84229 

37 

24 

•27545 

3.63048 

•29432 

3-39771 

•31338 

3.19100 

.33266 

3.00611 

•35216 

2.83965 

36 

25 

.27576 

3.62636 

.29463 

3-39406 

•31370 

3-18775 

•33298 

3.00319 

•35248 

2  .  83702 

35 

26 

.27607 

3.62224 

•29495 

3-39042 

.31402 

3-18451 

•3333° 

3.00028 

•3528r 

2.83439 

34 

27 

.27638 

3.61814 

.29526 

3-38679 

•31434 

3.18127 

•33363 

2.99738 

•353*4 

2.83176 

33 

28 

.27670 

3.61405 

•29558 

3-38317 

.31466 

3.17804 

•33395 

2-99447 

•35346 

2.82914 

32 

2Q 

.27701 

3.60996 

•29590 

3-37955 

.31498 

3.17481 

•33427 

2.99158 

•35379 

2.82653 

31 

30 

•27732 

3.60588 

.29621 

3-37594 

•31530 

3-i7i59 

•3346o 

2.98868 

•35412 

2.82391 

30 

31 

.27764 

3.60181 

•29653 

3-37234 

.31562 

3.16838 

•33492 

2.98580 

•35445 

2.82130 

29 

32 

•27795 

3-59775 

.29685 

3-36875 

•3I594 

3.16517 

•33524 

2.98292 

•35477 

2.81870 

28 

33 

.27826 

3  •  59373 

.29716 

3.36516 

.31626 

3.16197 

•33557 

2  .  98004 

•35510 

2.  8l6lO 

27 

34 

.27858 

3-58966 

•29748 

3-36158 

.31658 

3-I5877 

•33589 

2.97717 

•35543 

2.81350 

26 

35 

.27889 

3-58562 

.29780 

3-358oo 

.31690 

S-JSSS8 

.33621 

2.97430 

•35576 

2.81091 

25 

36 

.27921 

3.58160 

.29811 

3-35443 

.31722 

3.15240 

•33654 

2.97144 

.35608 

2-80833 

24 

37 

•27952 

3-57758 

•29843 

3-35087 

•31754 

3.14922 

.33686 

2.96858 

•35641 

2.80574 

23 

38 

•27983 

3-57357 

•29875 

3-34732 

.31786 

3.14605 

.33718 

2.96573 

.35674 

2.80316 

22 

39 

.28015 

3  •  56957 

.  29006 

3-34377 

.31818 

3.14288 

•33751 

2.96288 

.35707 

2  .  80059 

21 

40 

.28046 

3-56557 

.29938 

3-34023 

.31850 

3-13972 

•33783 

2  .  96004 

•35740 

2.79802 

2O 

4i 

.28077 

3-56l59 

.29970 

3-33670 

.31882 

3.13656 

.33816 

2.95721 

•35772 

2-79545 

J9 

42 

.28109 

3-5576r 

.30001 

3-333I7 

•3I9i4 

3-13341 

.33848 

2-95437 

•35805 

2.79289 

18 

43 

.28140 

3o5364 

.30033 

3-32965 

.31946 

3.13027 

•33881 

2.95155 

•35838 

2.79033 

*7 

44 

.28172 

3.54968 

.30065 

3.32614 

•3T978 

3-I27I3 

•339*3 

2.94872 

•35871 

2.78778 

16 

45 

.28203 

3-54573 

.30097 

3.32264 

.32010 

3.12400 

•33945 

2.94591 

•35904 

2    78523 

15 

46 

.28234 

3-54*79' 

.30128 

3-3I9I4 

.32042 

3.12087 

•33978 

2  .  94309 

•35937 

2  .  78269 

14 

47 

.28266 

3-53785 

.30160 

3-3I565 

.32074 

3-I1!775 

.34010 

2.94028 

•35969 

2.78014 

J3 

48 

.28297 

3-53393 

.30192 

3.31216 

.32106 

3.11464 

•34043 

2.93748 

.36002 

2.77761 

12 

49 

•28329 

3.53001 

.30224 

3.30868 

•32139 

3-1:[I53 

•34075 

2.93468 

•36035 

a-  775°7 

II 

5° 

.28360 

3.52609 

•30255 

3.30521 

.32171 

3.10842 

.34108 

2.93189 

.  36068 

2.77254 

IO 

5i 

.28391 

3.52219 

.30287 

3-30I74 

.32203 

3-10532 

.34140 

2.92910 

.36101 

2.77002 

9 

52 

.28423 

3.51829 

•3°3I9 

3.29829 

•32235 

3.10223 

•34173 

2.92632 

•36134 

2.76750 

8 

53 

•28454 

3-5I44t 

•30351 

3.29483 

.32267 

3.09914 

•34205 

2.92354 

.36167 

2.76498 

7 

54 

.28486 

3-51053 

.30382 

3-29I39 

.32299 

3.09606 

•34238 

2.92076 

.36199 

2.76247 

6 

55 

.28517 

3.50666 

.30414 

3-28795 

•32331 

3.09298 

•34270 

2.91799 

.36232 

2  .  75996 

5 

56 

•28549 

3.50279 

.30446 

3'  28452 

.32363 

3.08991 

.34303 

2.91523 

.36265 

2.75746 

4 

57 

.28580 

3.49894 

.30478 

3.28109 

.32396 

3-08685 

•34335 

2.91246 

.36298 

2.75496 

3 

58 

.28612 

3-49509 

•  3°5°9 

3.27767 

.32428 

3-08379 

-34368 

2.90971 

•36331 

2.75246 

2 

59 

.28643 

3.49125 

•30541 

3.27426 

.32460 

3.08073 

.34400 

2  .  90696 

•36364 

2.74997 

I 

60 

•28675 

3.48741 

•30573 

3.27085 

.32402 

3.07768 

•34433 

2.90421 

•36397 

2.74748 

O 

/ 

Cotang 

Tang 

Cotang 

Tang 

Cotang 

Tang 

Cotang 

Tang 

Cotang 

Tang 

r 

74° 

73° 

72° 

71° 

7°° 

90 


NATURAL  TANGENTS  AND  COTANGENTS. 


2 

0° 

2 

i° 

2 

2° 

2 

3° 

2 

4° 

Tang 

Cotang 

Tang 

Cotang 

Tang 

Cotang 

Tang 

Cotang 

Tang 

Cotang 

o 

•36397 

2.74748 

•38386 

2  .  60509 

.  40403 

2.47509 

•42447 

2-35585 

-44523 

.24604 

60 

I 

•36430 

2-74499 

.38420 

2.60283 

.40436 

2.47302 

.42482 

2-35395 

.44558 

.24428 

59 

2 

3 

•  36463 
.  36496 

2.74251 
2.74004 

•38453 
.38487 

2.60057 
2.59831 

.40470 
.  40504 

2.47095 
2.46888 

.42516 
•42551 

2  -  35205 
^•35oi5 

•44593 
.44627 

.24252 
.24077 

57 

4 

•  36529 

2-73756 

.38520 

2.59606 

.40538 

2.46682 

.42585 

2.34825 

.44662 

•  23902 

56 

5 

.36562 

2.73509 

•38553 

2.59381 

.40572 

2.46476 

.42619 

2  •  34636 

.44697 

.23727 

55 

6 

•36595 

2.73263 

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PHYSICS. 


MATTER  AOT>  ITS  PTtOPEHTIES. 


DEFINITIONS. 

1.  Natural    and    Physical    Science. — Science,   which 
may  be  denned  as  a  classified  knowledge  of  nature,  is  divided 
into  natural  and  physical  science. 

Natural  science  concerns  itself  with  the  external  form 
and  internal  structure  of  bodies. 

Physical  science  considers  only  the  matter  of  which 
these  bodies  are  composed. 

ILLUSTRATION. — Geology,  mineralogy,  botany,  and  zoology,  which 
investigate  the  form  and  structure  of  the  earth,  of  minerals,  of  plants, 
and  of  animals,  respectively,  are  natural  sciences. 

Physics  and  chemistry,  which  consider  the  properties  of  matter 
itself,  whether  it  is  light  or  heavy,  hard  or  soft,  combustible  or  non- 
combustible,  are  physical  sciences. 

2.  Matter  is  anything  that  possesses  weight;    that  is, 
is  acted  on  by  gravitation.      In  studying  matter,   physical 
science  considers: 

1.  The  division  of  which  matter  is  capable. 

2.  The  attractions  by  which    these    particles    are   held 
together. 

3.  The  motions  these  particles  may  have. 

3.  Division  of  Matter. — Science  recognizes  three  divi- 
sions of  matter:  masses,  molecules,  and  atoms. 

§  4. 


2  PHYSICS.  §  4 

A  mass,  or  body,  of  matter  is  any  portion  of  matter  appre- 
ciable by  the  senses. 

A  molecule  is  the  smallest  particle  of  matter  that  a  body 
can  be  divided  into  without  losing  its  identity. 

An  atom  is  an  indivisible  portion  of  matter. 

Atoms  unite  to  form  molecules;  a  collection  of  molecules 
forms  a  mass,  or  a  body. 

It  has  been  calculated  that  the  diameter  of  a  molecule  is 
larger  than  ^^rik^-^  of  an  inch,  and  smaller  than  ^^^^ 
of  an  inch. 

ILLUSTRATION. — The  sun  and  the  grain  of  sand  are  both  masses  of 
matter ;  the  smallest  particles  of  sugar  or  of  salt  that  still  show  the 
properties  of  these  substances,  respectively,  are  molecules  of  sugar  or 
of  salt.  The  still  more  minute  particles  of  carbon,  of  hydrogen,  and 
of  oxygen,  which  make  up  the  molecule  of  sugar,  or  those  of  chlorine 
and  of  sodium,  which  compose  the  molecule  of  salt,  are  atoms. 

4.  Attraction  of  Matter. — Three  forms  of  attraction 
are  recognized  in  science : 

1.  That  form  of  attraction  called  gravitation,  which  is 
exerted  between  masses  of  matter. 

2.  That   form  which    binds   molecules   together;    called 
cohesion,  when  the  molecules  are  alike,  and  adhesion,  when 
the  molecules  are  unlike. 

3.  That  form  of  attraction  called  chemical  attraction,  or 
affinity,  which  draws  atoms  together. 

ILLUSTRATION. — The  planets  are  held  to  the  sun,  and  stones  are  held 
to  the  earth  by  the  attraction  of  gravitation.  The  molecules  of  silver 
or  salt,  being  alike,  are  held  together  by  cohesion ;  those  of  granite  or 
gunpowder,  being  unlike,  are  held  together  by  adhesive  attraction. 
The  atoms  that  form  a  molecule  of  sugar  or  of  salt  are  united  by  chem- 
ical attraction. 

Molecules  united  by  cohesion,  therefore,  form  homogeneous  matter, 
while  unlike  molecules  united  by  adhesion  form  heterogeneous  matter. 
It  is  evident  that  there  must  be  as  many  kinds  of  molecules  as  there 
are  kinds  of  homogeneous  matter. 

5.  Motions  of  Matter. — The  forms  of  motion  admitted 
in  science  are  three : 

1.  Mass  motion,  or  visible  mechanical  motion. 

2.  Molecular  motion,  the  motion  of  the  molecule  within 


§  4  PHYSICS.  3 

the  mass,  called  heat,  light,  electricity,  or  magnetism,  accord- 
ing to  its  character. 

3.  Atomic  motion,  which,  though  probable,  is  not  yet 
fully  established. 

6.  Province  of  Physics. — Physics  is  that  department  of 
physical  science  which  studies  the  results  that  flow  from  the 
molecular  conditions  of  matter. 

ILLUSTRATION. — Weight,  which  is  a  consequence  of  mass  attraction; 
impact,  which  is  a  result  of  mass  motion;  tenacity,  hardness,  and 
elasticity,  which  depend  on  cohesion ;  solution,  capillarity,  and  diffusion, 
which  result  from  adhesion;  heat,  light,  electricity,  and  magnetism, 
which  are  molecular  motions — these  are  all  objects  of  physical  study 
and  investigation. 

7.  Bodies  are,  as  we  have  seen,  collections  of  molecules. 
They  exist  in  three  forms  or  conditions :  solid,  liquid,  and 
gaseotis. 

A  solid  body  is  one  whose  molecules  change  their  relative 
positions  with  great  difficulty;  as,  iron,  wood,  stone,  etc. 

A  liquid  body  is  one  whose  molecules  tend  to  change 
their  relative  position  easily.  Liquids  readily  adapt  them- 
selves to  the  vessel  that  contains  them,  and  their  upper  sur- 
face always  tends  to  become  level.  Water,  mercury,  etc. 
are  liquids. 

A  gaseous  body,  or  a  gas,  is  one  whose  molecules  tend  to 
separate  from  one  another;  as,  air,  oxygen,  etc. 

Gaseous  bodies  are  sometimes  called  aeriform  (air-like) 
bodies.  They  are  divided  into  two  classes :  permanent  gases 
and  vapors.  • 

A  permanent  gas  is  one  that  remains  a  Ngas  at  ordinary 
temperatures  and  pressures. 

A  vapor  is  a  body  that  at  ordinary  temperatures  is  a 
liquid  or  solid,  but  when  heat  is  applied  becomes  a  gas. 

One  body  may,  however,  exist  in  all  three  states;  as,  for 
example,  mercury,  which  at  ordinary  temperatures  is  a 
liquid,  becomes  a  solid  (freezes)  at  —40°  C.,  and  a  vapor  at 
350°  C.  Nearly  all  gases  can  be  liqiiefied  by  great  cold,  and 
some  can  even  be  solidified.  By  means  of  heat  nearly  all 
solids  may  be  either  liquefied  or  vaporized. 


4  PHYSICS.  §  4 

8.  Properties  of  Matter. — All  matter  possesses  Certain 
qualities  that  are,  in  scientific  language,  called  properties. 
Some  properties  are  common  to  all  kinds  of  matter ;  other 
properties  are  possessed  only  by  certain  kinds   of  matter. 
The  former,  termed  general  proper-ties,  are:  extension,  impene- 
trability, weight,  inertia,  mobility,  divisibility,  porosity,  com- 
pressibility,   expansibility,    elasticity,    and    indestructibility. 
Among  the  latter,  which  are  known  as  special  properties,  are 
the  following:    hardness,   tenacity,   brittleness,    malleability, 
and  ductility. 

9.  Extension  is  the  property  of  occupying  space.    Since 
all  bodies  must  occupy  space,  it  follows  that  extension  is  a 
general  property. 

10.  Impenetrability  is   the  property    of   matter   that 
prevents  two  bodies  from  occupying  the  same  portion  of 
space  at  the  same  time. 

1 1 .  Weight  is  the  measure  of  the  earth's  attraction  upon 
a  body.     All  bodies  have  weight.      In  former  times  it  was 
supposed  that  gases  had  no  weight,  since,  if  unconfined,  they 
tend  to  move  away  from  the  earth ;  but,  nevertheless,  they 
will  finally  reach  a  point  beyond  which  they  cannot  go,  being 
held   in   suspension   by  the   earth's  attraction.     Weight   is 
measured  by  comparison  with  some  standard. 

12.  Inertia  means  that  a  body  can  neither  put  itself  in 
motion  nor  bring  itself  to  rest.     To  do  so,  it  must  be  acted 
on  by  some  force. 

13.  Mobility   means   that   a  body  can  be   changed  in 
position  by  some  force  acting  on  it. 

14.  Divisibility  is  that  property  of  matter  which  indi- 
cates that  a  body  may  be  separated  into  parts. 

15.  Porosity  is  that  property  of  matter  which  indicates 
that  there  is  space  between  the  molecules  of  a  body.     Mole- 
cules of  bodies  are  supposed  to  be  spherical,  and  hence  there 
is  space  between  them,  as  there  would  be  between  peaches 
in  a  basket.    It  is  said  that  the  molecules  of  water  are  larger 


§  4  PHYSICS.  5 

than  those  of  salt;  so  that  when  salt  is  dissolved  in  water, 
its  molecules  wedge  themselves  between  the  molecules  of 
the  water,  and  unless  too  much  salt  is  added,  the  water  will 
occupy  no  more  space  than  it  did  before.  This  does  not 
prove  that  water  is  penetrable,  for  the  molecules  of  salt 
occupy  the  space  between  the  molecules  of  water. 

Water  has  been  forced  through  iron  by  pressure,   thus 
proving  that  iron  is  porous. 

16.  Compressibility  is  that  property  of  matter  which 
indicates  that  the  molecules  of  a  body  may  be  crowded  nearer 
together,  so  as  to  occupy  a  smaller  space. 

17.  Expansibility  is  that  property  of  matter  which  indi- 
cates that  the  molecules  of  a  body  may  be  forced  apart,  so 
as  to  occupy  a  greater  space. 

18.  Elasticity  is  that  property  of  matter  which  indi- 
cates that  if  a  body  be   distorted   within   certain  limits,  it 

-will  resume  its  original  form  when   the  distorting  force  is 
removed.     Glass,  ivory,  and  steel  are  more  or  less  elastic. 

19.  Indestructibility  indicates  that  matter  can  never 
be  destroyed.     A  body  may  undergo  thousands  of  changes; 
it  may  be  resolved  into  its  molecules,  and  its  molecules  may 
be  resolved  into  atoms;  these  atoms  may  unite  with  other 
atoms  to  form  other  molecules  and  bodies,  which  may  be 
entirely  different  from   the  original  body;  but  through  all 
these  changes  the  number  of  atoms  remains  the  same.     The 
whole  number  of  atoms  in  the  universe  is  exactly  the  same 
now  as  it  was  millions  of  years  ago,  and  will  always  remain 
the  same.     Matter  is  indestructible. 

20.  Hardness  is  that  property  of  matter  which  indi- 
cates that  some  bodies  may  scratch  other  bodies.      Fluids 
and  gases  do  not  possess   hardness.     The  diamond  is  the 
hardest  of  all  substances. 

21.  Tenacity  is  that  property  of  matter  which  indicates 
that  some  bodies  resist  a  force  tending  to  pull  them  apart. 
Steel  is  very  tenacious. 


6  PHYSICS.  §  4 

22.  Brittleness  is  that  property  of  matter  which  indicates 
that  some  bodies  are  easily  broken;  as,  glass,  crockery,  etc. 

23.  Malleability  is  that  property  of  matter  which  indi- 
cates that  some  bodies  may  be  hammered  or  rolled  into 
sheets.     Gold  is  the  most  malleable  of  all  substances. 

24.  Ductility  is  that  property  of  matter  which  indicates 
that  some  bodies  may  be  drawn  into  wire.      Platinum  is  the 
most  ductile  of  all  substances. 


GRAVITATION. 

25.  Every  body  in  the  universe  exerts  a  certain  attract- 
ive force  on  every  other  body,  which  tends  to  draw  the  two 
bodies  together.     This  attractive  force  is  called  gravitation. 

If  a  body  is  held  in  the  hand,  a  downward  pull  is  felt,  and 
if  released,  it  will  fall  to  the  ground."" This"piuTis  commonly 
called  weight,  but  it  really  is  the  attraction  between  the 
earth  and  the  body. 

26.  Force   of  gravity   is   a   term  used  to  denote  the 
attraction  between   the  earth  and  bodies  upon  or  near  its 
surface.    It  always  acts  in  a  straight  line  between  the  center 
of  the  body  and  the  center  of  the  earth. 

The  force  of  gravity  varies  at  points  on  the  earth's  sur- 
face. It  is  slightly  less  on  the  top  of  a  high  mountain  than 
at  the  level  of  the  sea.  For  this  reason,  the  weight  of  a 
body  also  varies.  But  if  the  weight  of  a  body  at  any  place 
be  divided  by  the  force  of  gravity  at  that  place,  the  result  is 
called  the  mass  of  the  body. 

'27.     The  mass  of  a  body  is  the  measure  of  the  actual 
amount  of  matter  that  it  contains,  and  is  ahvays  the  same. 
Let          m  —  mass  of  the  body; 
W  —  weight  of  the  body; 

g    —  force  of  gravity  at  the  place  where  the  body 
was  weighed. 

weigfht  of  bodv  W 

Then,       mass  =  - — $ — ^Z    Or  m  —  — .         (1.) 

force  or  gravity  g 


§  4  PHYSICS.  7 

28.  Law  of  Gravitation. —  The  force  of  attraction   by 
which  one  body  tends  to  draw  another  body  towards  it  is 
directly  proportional  to  its  mass,  and  inversely  proportional  to 
the  square  of  the  distance  between  the  centers  of  the  two  bodies. 

29.  Laws  of  Weight. — Bodies  weigh  most  at  the  surface 
of  the  earth.     Below  the  surface,  the  weight  decreases  as  the 
distance  to  the  center  decreases. 

A  bove  the  surface  the  weight  decreases  as  the  square  of  the 
distance  increases. 

ILLUSTRATION. — If  the  earth's  radius  is  4,000  miles,  a  body  that 
weighs  100  pounds  at  the  surface  will  weigh  nothing  at  the  center, 
since  it  is  attracted  in  every  direction  with  equal  force.  At  1,000  miles 
from  the  center  it  will  weigh  25  pounds,  since 

4,000  :  1,000  =  100  :  25. 
At  2,000  miles  from  the  center  it  will  weigh  50  pounds,  since 

4,000  :  2,000  =  100  :  50. 

At  3,000  miles  from  the  center  it  will  weigh  75  pounds,  and  at  the 
surface,  or  4,000  miles  from  the  center,  it  will  weigh  100  pounds.  If 
carried  still  higher,  say  1,000  miles  from  the  surface,  or  5,000  miles  from 
the  center  of  the  earth,  it  will  weigh  64  pounds,  since 

5,0002  :  4,000'2  =  100  :  64. 

At  4,000  miles  from  the  surface  it  will  weigh  25  pounds,  since 
8,0002  :  4,000'  =  100  :  25. 

30.  Formulas  for  Gravity  Problems. 

Let     W  =  weight  of  a  body  at  the  surface; 

w  =  weight  of  a  body  at  a  given  distance  above  or 

-   below  the  surface; 
d  —  distance  between  the  center  of  the  earth  and 

the  center  of  the  body ; 
R  —  radius  of  the  earth  =  4,000  miles. 

Formula  for  weight  when  the  body  is  below  the  surface, 

wR-dW.  (2.) 

Formula  for  weight  when  the  body  is  above  the  surface, 

wd*  =   WR\          (3.) 

EXAMPLE  1. — How  far  below  the  surface  of  the  earth  will  a  25-pound 
ball  weigh  9  pounds  ? 


8  PHYSICS.  §  4 

SOLUTION. — Use  formula  3,  w  R  =  dW. 
Substituting  the  values  of  /?,  W,  and  w,  we  have 
9X4,000  =  </x25, 

n \/  A   QAf) 

or  d  =  -^ =  1,440  miles  from  the  center. 

60 

4,000  -  1,440  =  2,560  miles  below  the  surface.     Ans. 

EXAMPLE  2. — If  a  body  weighs  700  pounds  at  the  surface  of  the  earth, 
at  what  distance  above  the  earth's  surface  will  it  weigh  112  pounds  ? 
SOLUTION.— Use  formula  3,  iu  d*  —  WR*. 
Substituting  the  values  of  R,  IV,  and  w,  we  have 
112  X  d*  —  700  X  4,000s, 


-V* 


'TOO  x  4.000;  orloowmiles< 

112 

Therefore,  10,000  —  4,000  —  6,000  miles   above   the   earth's  surface. 

Ans. 


EXAMPLE  3.— The  top  of  Mt.  Hercules  is  said  to  be  32,000  feet,  say 
6  miles,  above  the  level  of  the  sea.  If  a  body  weighs  1,000  pounds  at 
sea  level,  what  would  it  weigh  if  carried  to  the  top  of  the  mountain  ? 

SOLUTION.—    w  d'2  =  W R*,  or  w  X  4,0062  =  1,000  X  4,0002 ; 

whence,  w  =  4'00^6|'00°  =  997  pounds.     Ans.    - 


EXAMPLES  FOR  PRACTICE. 

31.      Solve  the  following : 

1.  How  much  would  1,000  tons  of  coal  weigh  1  mile  below  the  sur- 
face ?  Ans.  1,999, 500  Ib. 

2.  How  much  would  the  coal  in  example  1  weigh  1  mile  above  the 
surface?  Ans.  1, 999, 000 Ib.,  nearly. 

3.  How  far  above  the  earth's  surf  ace  would  it  be  necessary  to  carry 
a  body  in  order  that  it  may  weigh  only  half  as  much  ? 

Ans.   1,656.854  mi.,  nearly. 

4.  A  man  weighs  160  pounds  at  the  surface;  how  much  will  he 
weigh  50  miles  below  the  surface  ?  Ans.   158  Ib. 

5.  If  a  body  weighs  100  pounds  400  miles  above  the  earth's  surface, 
how  much  will  it  weigh  at  the  surface  ?  Ans.  121  Ib. 

NOTE. — Use  4,000  miles  as  the  radius  of  the  earth. 


32.     The  specific  gravity  of  a  body  &  the  ratio  between 
its  weight  and  the  weight  of  a  like  volume  of  water. 


§  4  PHYSICS.  9 

Since  gases  are  so  much  lighter  than  water,  it  is  usual  to 
take  the  specific  gravity  of  a  gas  as  the  ratio  between  the 
weight  of  a  certain  volume  of  the  gas  and  the  weight  of  the 
same  volume  of  air. 

EXAMPLE. — A  cubic  foot  of  cast  iron  weighs  450  pounds;  what  is  its 
specific  gravity,  a  cubic  foot  of  water  weighing  62.42  pounds  ? 
SOLUTION. — According  to  the  definition,  the  specific  gravity  is 
450 


62.42 


=  7.21.     Ans. 


33.  The  specific  gravities  of  different  bodies  are  given  in 
printed  tables ;  hence,  if  it  is  desired  to  know  the  weight  of 
a  body  that  cannot  be  conveniently  weighed,  calculate  its 
cubical  contents  and  multiply  tJie  specific  gravity  of  the  body 
by  the  weight  of  a  like  volume  of  water,  remembering  that  a 
cubic  foot  of  water  weighs  62.  J$  pounds. 

EXAMPLE. — How  much  will  3,214  cubic  inches  of  cast  iron  weigh  ? 
Take  its  specific  gravity  as  7.21. 

SOLUTION. — Since  1  cubic  foot  of  water  weighs  62.42  pounds,  3,214 
cubic  incnes  weigh 

•^|i|  X  62.42  =  116.098  pounds. 

1 ,   I  /wO 

Then,  116.098  X  7.21  =  837.067  pounds.     Ans. 


EXAMPLES  FOR  PRACTICE. 

34.  Solve  the  following: 

1.  If  a  cubic  foot  of  a  certain  substance  weighs  162|  pounds,  what 
is  its  specific  gravity  ?  Ans.  2.6. 

2.  A  mixture  of  lead  and  tin  has  a  specific  gravity  of  9.28;  what  is 
the  weight  of  a  cubic  inch  ?  Ans.  5.36  oz. 

3.  What  is  the  weight  of  a  cubic  foot  of  anthracite  coal  having  a 
specific  gravity  of  1.55  ?  Ans.  96.751  Ib. 

4.  Find  the  weight  of  100  cubic  yards  of  earth  having  a  specific 
gravity  of  1.75.  Ans.  147.467  T. 

35.  The  following  tables  give  the  specific  gravities  of  a 
variety  of  substances  likely  to  be  met  with  in  ordinary  prac- 
tice.   The  weights  per  cubic  foot  are  calculated  on  a  basis  of 
62.5  pounds  of  water  per  cubic  foot 


10 


PHYSICS. 
TABLE   1. 


APPROXIMATE  SPECIFIC  GRAVITIES  OF  VARIOUS 
SUBSTANCES. 


METALS. 


Substance. 

Specific 
Gravity. 

Weight  per 
Cubic  Foot 
in  Pounds. 

Osmium  

22   48 

1  405.0 

Platinum 

21   50 

1  343  8 

Gold  

19  .  50 

1,218.8 

Mercury  

13  60 

850  0 

Lead  (cast)  

11  35 

709  4 

Silver  

10  50 

656  3 

Copper  feast)  .  . 

8  79 

549  4 

Brass  

8  38 

523  8 

Wrought  iron  

7.68 

480  0 

Cast  iron 

7  21 

4t>0  0 

Steel  

7.84 

490.0 

Tin  (cast) 

7  29 

455  6 

Zinc  (cast)           

6  86 

428  8 

Antimony         

6  71 

419  4 

Aluminum 

2  60 

162  5 

LIQUIDS. 


Substance. 

Specific 
Gravity. 

Weight  per 
Cubic  Foot 
in  Pounds. 

Ascetic  acid 

1   062 

66  4 

Nitric  acid  

1.420 

88.8 

Sulphuric  acid  

1.841 

115.1 

Muriatic  acid  

1.200 

75.0 

Alcohol 

800 

50  0 

Turpentine 

870 

54  4 

Sea  -water  (ordinary) 

1  026 

64  1 

Milk  

1  032 

64  5 

PHYSICS. 

WOODS. 


11 


Substance. 

Specific 
Gravity. 

Weight  per 
Cubic  Foot 
in  Pounds. 

Ash  .-  

845 

52   80 

Beech 

852 

53  25  • 

Cedar 

561 

35  06 

Cork   .    . 

240 

15  00 

Ebony  (American) 

1  331 

83  19 

Lignum  vitse 

1  333 

83  30 

Maple           .  . 

750 

46  88 

Oak  (old)  

1.170 

73  10 

Spruce  

.500 

31.25 

Pine  (yellow)  

.660 

41  .  20 

Pine  (white) 

554 

34  60 

Walnut 

671 

41   90 

At 


GASES. 
F.,  and  Under  a  Pressure  of  One  Atmosphere. 


Substance. 

Specific 
Gravity. 

Weight  per 
Cubic  Foot 
in  Pounds. 

Atmospheric  air 

1   0000 

08073 

Carbonic  acid 

1   5290 

12344 

Carbonic  oxide 

9674 

07810 

Chlorine 

2  4400 

19700 

Oxvp^en 

1  1056 

08925 

Nitrogen  

9713 

07841 

Smoke  (bituminous  coal)  

1020 

00815 

Smoke  (wood)  

0900 

00727 

*Steam  at  212°  F  

.4700 

.03790 

Hydrogen  .        .                         . 

.0692 

00559 

*The  specific  gravity  of  steam  at  any  temperature  and  pressure, 
compared  with  air  at  the  same  temperature  and  pressure,  is  0.622. 


PHYSICS. 


MISCELLANEOUS. 


Substance. 

Specific 
Gravity. 

Weight  per 
Cubic  Foot 
in  Pounds. 

Kmery  

4.00 

250 

Glass  (average) 

2  80 

175 

Chalk         ... 

2  78 

174 

Granite  ... 

2  65 

166 

Marble   

2  70 

169 

Stone  (common)  

2  52 

158 

Salt  (common) 

2  13 

133 

Soil  (common) 

1  98 

124 

Clay 

1  93 

121 

Brick  

1  90 

118 

Plaster  of  Paris  (average) 

2  00 

125 

Sand  

1.80 

113 

36.     Determining  Specific  Gravity  of  Liquids. — To 

find  the  specific  gravity  of  any  particular  liquid,  compared 
with  that  of  water,  it  is  only  necessary  to  weigh  equal  bulks 
and  divide  the  weight  of  the  liquid  by  the  weight  of  the 
water.  To  weigh  eqtial  bulks  of  liquid  the 
simplest  and  most  accurate  way  is  to  weigh 
them  in  succession  in  the  same  vessel,  taking 
care  that  it  is  equally  full  on  both  occasions. 

A  convenient  vessel,  commonly  used  when 
small  quantities  are  to  be  weighed,  is  shown  in 
Fig.  1.  It  is  easily  made  by  blowing  a  bulb  on 
a  glass  tube.  On  that  portion  of  the  tube  which 
is  narrowed  by  drawing  it  out  over  a  flame,  a 
scratch  is. made  with  a  diamond  or  a  file.  The 
bulb  is  filled  up  to  the  scratch  with  the  liquid, 
and  is  then  weighed,  emptied,  cleaned,  dried, 
filled  with  water,  and  again  weighed.  In  these 
experiments  care  should  always  be  taken  that  both  liquids 
have  the  same  temperature.  This  is  easily  accomplished  by 


FIG.  l. 


PHYSICvS. 


13 


immersing  the  bulb  filled  with  liquid  for  some  time  in  water, 
part  of  which  is  later  used  in  the  second  weighing. 

Another  form  of  apparatus  for  determining  the  specific 
gravity  of  liquids,  devised  by  Dr.  Sprengel,  consists  of  an 
elongated  U  tube,  Fig.  2,  the  ends  of  which  terminate  in  the 
two  capillary  tubes  A,  B,  bent  at  right  angles  in  opposite 
directions.  The  shorter  one  A  is  narrower  at  the  end  than 
the  longer  one.  The  horizontal  part  of  the  wider  tube  is 
marked  near  the  bend  with  a  fine  line.  The  U  tube  is  filled 
by  suction,  the  little  bulb  apparatus,  Fig.  3,  having  been 
previously  attached  to  the  narrow  capillary  tube  by  a  piece 
of  rubber  tubing.  It  is  then  detached  from  the  bulb,  placed 


FIG.  2. 

in  water  almost  to  the  bends  of  the  capillary  tubes,  left  there 
until  it  has  assumed  the  temperature  of  the  water,  and  after 
careful  adjustment  of  the  volume  of  the  liquid  up  to  the 
mark  in  the  wider  capillary  tube,  it  is  taken  out,  dried,  and 
weighed. 

37.     Determining  the  Specific  Gravity  of  Solids.— 

The  determination  of  the  specific  gravity  of  a  solid  body  is 
also  made  according  to  the  principles  explained,  and  may  be 
made  with  the  specific-gravity  bottle.  The  bottle  is  first 
weighed  full  of  water;  the  solid  is  then  placed  in  the  same 
pan  of  the  balance  and  its  weight  determined;  finally,  the 
solid  is  put  into  the  bottle,  displacing  an  equal  bulk  of  water, 
the  weight  of  which  is  determined  by  the  loss  on  again  weigh- 
ing. In  this  way,  the  weights  of  the  solid  and  that  of  an 


14 


PHYSICS. 


equal  bulk  of  water  are  obtained.     The  former  divided  by 
the  latter  gives  the  specific  gravity. 

EXAMPLE. — A  piece  of  lead  weighs 73.68  grains. 

Glass  bottle  filled  with  water  weighs 297.05  grains. 

Total 370.73  grains. 

After  an  equal  volume  of  water  was  dis- 
placed by  the  lead,  the  weight  was 864.15  grains. 

Hence,  the  displaced  water  weighed      6.58  grains. 

r/o   ^*o 

Therefore,  the  specific  gravity  of  lead  is  ~  ^-  —  11.2—.     Ans. 

o.  oo 

38.  Another  ingenious,  but  not  quite  as  exact,  method 
of  finding  the  specific  gravity  of  solids  not  easily  soluble  in 
water  is  based  on  the  principle  of  Archimedes: 

When  a  body  is  im- 
mersed in  a  fluid,  it  loses 
in  weight  an  amount  equal 
to  the  weight  of  the  fluid 
it  displaces.  , 

This  principle  is  ap- 
plied as  follows:  Weigh 
the  body  first  in  the  air, 
then  weigh  the  body  in 
water,  suspending  it  by 
a  string  attached  to  a 
scale  pan,  as  shown  in 
Fig.  4.  The  difference 
between  the  t^vo  weights 
will  be  the  weight  of  an 
FlG-  4-  equal  volume  of  water. 

The  ratio  of  the  weight  in  air  to  the  difference  thus  found 
will  be  the  specific  gravity. 

Let  Sp.  Gr.  =  specific  gravity ; 

W    —  weight  of  the  solid  in  air ; 
W   =  weight  of  the  solid  in  water. 

Then  the  weight  of  a  volume  of  water  equal  to  the  volume 
of  the  solid  is  W-  W,  and 

S-  Gr-  = 


§  4  PHYSICS.  15 

EXAMPLE.  —  A  body  in  air  weighs  36.25  grains,  and  in  water,  30  grains  ; 
what  is  its  specific  gravity  ? 
SOLUTION.  —  Using  formula  4,  we  obtain 

W  36.25  36.25 

SP-  Gr'  =    W-W    =  36.25-30  =  -6725   =  5'8'     AnS' 

39.  If  the  body  is  lighter  than  water,  a  piece  of  iron  or 
other  heavy  substance  must  be  attached  to  it,  sufficiently 
heavy  to  sink  both.  Then  weigh  both  bodies  in  air  and  both 
in  water.  Weigli  both  separately  in  air,  and  weigh  the  heavier 
body  in  water.  Subtract  the  weight  of  the  bodies  in  water 
from  their  iveight  in  air,  and  the  result  ivill  be  the  weight 
of  a  volume  of  the  water  equal  to  the  volume  of  the  two  bodies. 
Find  the  difference  of  the  weights  of  the  heavy  body  in  air 
and  in  water,  and  the  result  will  be  the  weight  of  a  volume 
of  water  equal  to  the  volume  of  the  heavy  body.  Subtract 
this  last  result  from  the  former,  and  the  result  will  be  the 
weight  of  a  volume  of  water  equal  to  the  volume  of  the  light 
body.  The  weight  of  the  light  body  in  air  divided  by  the 
weight  of  its  equal  volume  of  water  is  the  specific  gravity  of 
the  light  body. 

Let      W    —  weight  of  both  bodies  in  air  ; 

W  =  weight  of  both  bodies  in  water; 
w     =  weight  of  light  body  in  air  ; 
Wl    =  weight  of  heavy  body  in  air  ; 
W^    —  weight  of  heavy  body  in  water. 

Then,  the  specific  gravity  of  the  light  body  is  given  by 

SP-  Gr-  =  '  (5>) 


EXAMPLE.  —  A  piece  of  cork  weighs  4.8  grains  in  air.  A  piece  of  cast 
iron  weighs  36  grains  in  air  and  31  grains  in  water.  The  weight  of  the 
iron  and  cork  together  in  water  is  15  grains;  what  is  the  specific 
gravity  of  the  cork  ?  of  the  cast  iron  ? 

SOLUTION.  —  Substituting  in  formula  5  the  values  given, 

q      r  _  w  __   _  _  4.8  _ 

-  (  w-  W'}  -(Wi—  WJ)  ~  (40.8  —  15)  -  (36  -  31) 

=  0-  23,  the  specific  gravity  of  the  cork.     Ans. 


By  formula  4, 
Sp.  Gr.  —  -yr^  —  rr^j-  =  o£  —  or  =  7.2,  specific  gravity  of  the  iron.   Ans. 

¥¥      -    YV  OO    -   OJ. 


1G  PHYSICS.  §  4 

40.  Determining   Specific   Gravity  of  Gases. — The 

method  of  finding  the  specific  gravity  of  a  gas  is  of  great 
importance  to  the  chemist.  The  theory  of  the  operation  is 
as  simple  as  if  liquids  are  concerned,  but  as  the  process  is 
more  delicate  and  involves  certain  corrections  for  differences 
of  temperature  and  pressure,  it  appears  preferable  to  defer 
the  consideration  of  this  matter  for  the  present. 

4 1 .  Hydrometers. — Instruments  called  hydrometers  are 
in  general  use  for  determining  quickly  and  accurately  the 
specific  gravities  of  liquids  and  some  forms  of  solids.     There 
are  two  kinds ;  viz. , 

1.  Hydrometers  of  constant  weight,  as  Beaume's. 

2.  Hydrometers  of  constant  volume,  as  Nicholson's. 

42.  A  hydrometer  of  constant  weight  is  shown  in  Fig.  5. 
It  consists  of  a  glass  tube,  near  the  bottom  of  which  are  two 
bulbs.     The  lower  and  smaller  bulb  is  loaded  with  mercury 

or  shot,  so  as  to  cause  the  instrument  to  remain  in 
a  vertical  position  when  placed  in  the  liquid.  The 
upper  bulb  is  filled  with  air,  and  its  volume  is  such 
that  the  whole  instrument  is  lighter  than  an  equal 
volume  of  water. 

The  point  to  which  the  hydrometer  sinks  when 
placed  in  water  is  usually  marked,  the  tube  being 
graduated   above   and  below   in   such   a   manner 
that  the  specific  gravity  of  the  liquid  can  be  read 
directly.     It  is  customary  to  have  two  instruments : 
one  with  the  zero  point  near  the  top  of  the  stem, 
FIG.  s.       for  use  in  liquids  heavier  than  water,  and  the  other 
with  the  zero  point  near  the  bulb,  for  use  in  liquids  lighter 
than  water. 

These  instruments  are  more  commonly  used  for  determin- 
ing the  degree  of  concentration  or  dilution  of  certain  liquids, 
as  acids,  alcohol,  milk,  solutions  of  sugar,  etc.,  rather  than 
their  actual  specific  gravities.  They  are  then  known  as 
acidimeters,  alcoholometers,  lactometers,  saccharimeters,  etc., 
according  to  the  use  to  which  they  are  put. 


PHYSICS. 


17 


43.  Nicholson's  Hydrometer.  —  This  instrument  is 
shown  in  Fig.  6.  It  consists  of  a  hollow  cylinder  carrying  at 
its  lower  end  a  basket  d  heavy  enough  to  keep  the  apparatus 
upright  when  placed 
in  water.  At  the 
top  of  the  cylinder 
is  a  vertical  rod,  to 
which  is  attached  a 
shallow  pan  a  for 
holding  weights, 
etc.  The  cylinder 
must  be  of  such  size 
that  the  apparatus 
may  be  so  much 
lighter  than  water 
that  a  certain  weight 

W  must  be  placed  FIG.  6. 

in  the  pan  to  sink  it  to  a  given  point  c  on  the  rod.  The  body 
whose  specific  gravity  it  is  desired  to  find  must  weigh  less 
than  W.  It  is  placed  in  the  pan  a  and  enough  weight  w  is 
added  to  sink  the  point  c  to  the  water  level.  It  is  evident 
that  the  weight  of  the  given  body  is  W  —  w.  The  body  is 
now  removed  from  the  pan  a  and  placed  in  the  basket  d, 
an  additional  weight  being  added  to  sink  the  point  c  to  the 
water  level.  Represent  the  weight  now  in  the  pan  by  W  . 
The  difference  W  —w  is  the  weight  of  a  volume  of  water 
equal  to  the  volume  of  the  body.  Hence, 


EXAMPLE.  —  The  weight  necessary  to  sink  the  hydrometer  to  the 
point  c  is  16  ounces  ;  the  weight  necessary  when  the  body  is  in  the  pan 
a  is  7.3  ounces,  and  when  the  body  is  in  the  basket  d,  10  ounces;  what 
is  the  specific  gravity  of  the  body  ? 


SOLUTION. — Sp.  Gr.  = 


IV  -w 


16-7.3 


W'-w  ~  10-7.3 


_  8.7  _ 

~          -  *'    '  S< 


44.  Archimedes'  principle  gives  a  very  easy  and  accurate 
method  of  finding  the  volume  of  an  irregularly  shaped  body. 
Thus,  subtract  its  weight  in  water  from  its  weight  in  air ; 


18 


PHYSICS. 


divide  by  .03617  and  the  result  will  be  the  volume  in  cubic 
inches;  or  divide  by  62.5  and  the  result  will  be  the  volume 
in  cubic  feet. 

If  the  specific  gravity  of  the  body  is  known,  its  cubical 
contents  can  be  found  by  dividing"  its  weight  by  its  specific 
gravity,  and  then  dividing  again  by  either  .03617  or  62.5. 

EXAMPLE. — A  certain  body  has  a  specific  gravity  of  4.88  and  weighs 
76  pounds.     How  many  cubic  inches  are  there  in  the  body  ? 
76 


SOLUTION. — 


=  479.72cu.  in.     Ans. 


4. 38  X- 03617 

45.  "Since  the  weight  of  a  cubic  foot  of  water  varies  for 
different  temperatures,  and  with  the  amount  of  impurities  it 
contains,  it  is  necessary  to  have  some  standard  when  getting 
the  specific  gravity.  This  standard  is  pure  distilled  water  at 
its  maximum  density,  which  occurs  at  a  temperature  of 
39. 2°  F.  At  this  temperature  water  weighs  62. 425  pounds 
per  cubic  foot ;  but  for  ordinary  calculations  it  is  customary 
to  take  it  as  weighing  1,000  ounces,  or  62.5  pounds  per  cubic 
foot. 


PROPERTIES    OF    AIR    AND    GASES. 


TENSION  OF  GASES. 

46.     The  most  striking  feature  concerning  gases  is  that, 

no  matter  how  small  the  quan- 
tity may  be,  they  will  always 
jill  the  vessels  that  contain 
them.  If  a  bladder  or  football 
is  partly  filled  with  air  and 
placed  under  a  glass  jar  (called 
a  receiver),  from  which  the  air 
has  been  exhausted,  the  bladder 
or  football  will  immediately 
expand,  as  shown  in  Fig.  7. 
The  force  that  a  gas  always 
exerts  when  confined  in  a  lim- 
ited space  is  called  tension. 
FIG.  7.  The  word  "tension"  in  this 


§  4  PHYSICS.  19 

case  means  pressure,  and  is  only  used  in  this  sense  in  refer- 
ence to  gases. 

47.  As  water  is  the  most  common  liquid,  so  is  air  the 
most  common  gas.      It  was  supposed  by  the  ancients  that  air 
was  imponderable ;  i.e.,  that  it  had  no  weight,  and  not  until 
the  year  1650  was  the  contrary  proved.     A  cubic  inch  of  air, 
under  ordinary  conditions,   weighs  nearly  .31  grain.     The 
ratio  of  the  weight  of  air  to  water  is  about  1  :  774,  which 
means  that  air  is  only  T^¥  as  heavy  as  water.      It  is  a  well 
known  fact  that  if  a  body  is  immersed  in  water  and  weighs 
less  than  the  volume  of  water  it  displaces,  this  body  will  rise, 
extend  partly  but  of  the  water,  and  float.     The  same  is  true, 
to  a  certain  extent,  of  air.     If  a  vessel  made  of  light  material 
is  filled  with  a  gas  lighter  than  air,  so  that  the  total  weight 
of  both,  vessel  and  gas,  is  less  than  the  weight  of  the  volume 
of  air  that  they  displace,  the  vessel  will  rise ;  it    is  on   this 
principle  that  balloons  are  made. 

48.  Since  the  air   has  weight,   it   is   evident   that   the 
enormous   quantity  of  air  that  constitutes  the  •  atmosphere 
must  exert  a  considerable  pressure  upon  the  earth.     We  can 
easily  notice  this  atmospheric  pressure  upon  our  persons.     If 
we  place  the  hand  over  the  open  mouth  of  the  receiver  of  a 
powerful  air  pump,  we  find  that  the  hand  is  pressed  down- 
wards with  great  force.     The  total  atmospheric  pressure  that 
the  human  body  has  to  support  amounts  to  several  tons,  but 
this    pressure    is    not    felt    under    ordinary   circumstances 
because  it  is  exerted  equally  in  every  direction. 

49.  This  pressure  of  air  may  easily  be  proved  by  taking 
a  long  glass  tube,  closed  at  one  end,  and  filling  it  with  mer- 
cury.    If  the  finger  is  placed  over  the  open  end,  so  as  to 
keep  the  mercury  from  running  out,  and  the  tube  is  inverted 
and  placed  in  a  cup  of  mercury,  as  shown  in  Fig.  8,  the 
mercury  will  fall,  then  rise,  and  after  a  few  oscillations  will 
come  to  rest  at  a  height  above  the  top  of  the  mercury  in  the 
cup  equal   to  about  30  inches.     This  height  will  always  be 
constant  under  similar  atmospheric  conditions.     Now,  if  the 


20 


PHYSICS. 


atmosphere  has  weight,  it  must  press  upon  the  upper  sur- 
face of  the  mercury  in  the  glass  with  equal  intensity  upon 
every  square  unit,  except  upon  that  part  of  the  surface  occu- 
pied by  the  tube.  According  to  Pascal's  law,  the  pres- 
sure per  unit  of  area  exerted  anywhere  upon  a  mass  of  liquid 
is  transmitted  undiminished  in  all  directions,  and  acts  with 
the  same  force  upon  all  surfaces  in  a 
direction  at  right  angles  to  those  sur- 
faces. Therefore  the  pressure  of  the 
atmosphere  is  transmitted  in  all 
directions.  There  being  nothing  in 
the  tube  except  mercury  to  counter- 
balance the  upward  pressure  of  the 
air,  the  mercury  falls  in  the  tube 
until  it  exerts  a  downward  pressure 
on  the  upper  surface  of  the  mercury 
in  the  cup  sufficiently  great  to  coun- 
terbalance the  upward  pressure  pro- 
duced by  the  atmosphere.  In  order 
that  there  shall  be  equilibrium,  the 
pressure  of  the  air  per  unit  of  area 
on  the  upper  surface  of  the  mercury 
in  the  glass  must  equal  the  pressure 
(weight)  exerted  per  unit  of  area  by 
the  mercury  inside  of  the  tube.  Sup- 
pose the  area  of  the  inside  of  the 
tube  is  1  square  inch;  then,  since 
, mercury  is  13.6  times  as  heavy  as 
water,  the  weight  of  the  mercurial 
columnis. 03617X13.6X30  =  14.7574 
pounds.  The  actual  height  of  the 
mercury  is  a  little  less  than  30  inches,  and  the  actual 
weight  of  a  cubic  inch  of  distilled  water  is  a  little  less  than 
.03617  pound.  Taking  these  considerations  into  account, 
the  average  weight  of  the  mercurial  column  at  the  level  of 
the  sea  is  14.69  pounds,  or,  as  it  is  usually  expressed,  14.7 
pounds.  Since  this  weight,  exerted  upon  1  square  inch  of 
the  liquid  in  the  glass,  just  produced  equilibrium,  it  is  plain 


FIG.  8. 


PHYSICS. 


21 


that  the  pressure  of  the  outside  air  is  14.7  pounds  upon  every 
square  inch  of  surface. 

5O.  Vacuum. — The  space  between  the  upper  end  of  the 
tube  and  the  upper  surface  of  the  mercury  is  called  a  vacuum, 
meaning-  that  it  is  an  entirely  empty  space,  and  does  not 
contain  any  substance — solid,  liquid,  or  gaseous.  If.it  con- 
tained a  gas  of  some  kind,  no  matter  how  small 
the  quantity  might  be,  it  would  expand,  filling 
the  space,  and  its  tension  would  cause  the  column 
of  mercury  to  fall  and  become  shorter,  according 
to  the  amount  ot  gas  present.  The  space  is  then 
called  a  partial  vacuum.  If  the  mercury  falls 
1  inch,  so  that  the  column  is  only  29  inches  high, 
we  say,  in  ordinary  language,  that  there  are 
29  inches  of  vacuum.  If  it  falls  8  inches,  we  say 
that  there  are  22  inches  of  vacuum;  if  it  falls 
16  inches,  we  say  that  there  are  14  inches  of 
vacuum ;  etc.  Hence,  when  the  vacuum  gauge  of 
a  condensing  engine  shows  26  inches  of  vacuum, 
there  is  enough  air  in  the  condenser  to  produce  a 
pressure  of 

QQ  Og  A 

-X14.7  =  ^X14.7  =  1.96  Ib.  per  sq.  in. 
oU  oU 

In  all  cases  where  the  mercurial  column  is  used 
to  measure  a  vacuum,  the  height  of  the  column 
in  inches  gives  the  number  of  inches  of  vacuum. 
Thus,  if  the  column  were  5  inches  high,  or  the 
vacuum  gauge  showed  5  inches,  the  vacuum 
would  be  5  inches. 

If  the  tube  had  been  filled  with  water  instead  of 
mercury,  the  height  of  the  column  of  water  to 
balance  the  pressure  of  the  atmosphere  would 
have  been  30x13.6  =  408  inches  =  34  feet. 
This  means  that  if  a  tube  were  filled  with  water, 
inverted,  and  placed  in  a  dish  of  water  in  a  man- 
ner similar  to  the  experiment  made  with  the  mer- 
cury, the  resulting  height  of  the  column  of  water 
FIG.  9.  would  be  34  feet. 


22  PHYSICS.  §  4 

51.  The  Barometer. — The  barometer  is  an  instrument 
for  measuring  the  pressure  of  the  atmosphere.  There  are 
two  kinds  in  general  use — the  mercurial  and  the  aneroid 
barometer.  The  mercurial  barometer  is  shown  in  Fig.  9. 
The  principle  is  the  same  as  in  the  case  of  the  inverted  tube 
shown  in  Fig.  8.  The  tube  and  cup  at  the  bottom  are  pro- 
tected by  a  brass  or  iron  casing.  At  the  top  of  the  tube  is 
a  graduated  scale  that  can  be  read  to  TTVir  of  an  inch  by 


FIG.  10. 

means  of  a  vernier.  Attached  to  the  casing  is  an  accurate 
thermometer  for  determining  the  temperature  of  the  outside 
air  at  the  time  the  barometric  observation  is  taken.  This  is 
necessary,  since  mercury  expands  when  the  temperature  is 
increased,  and  contracts  when  the  temperature  falls;  for  this 
reason  a  standard  temperature  is  assumed,  and  all  barometer 
readings  are  reduced  to  this  temperature.  This  standard 


§  4  PHYSICS.  23 

temperature  is  usually  taken  at  32°  F.,  at  which  temperature 
the  height  of  the  mercurial  column  is  30  inches.  Another 
correction  is  made  for  the  altitude  of  the  place  above  sea 
level,  and  a  third  correction  for  the  effects  of  capillary 
attraction. 

52.  In  Fig.  10  is  shown  an  illustration  of   an  aneroid 
barometer.     These  instruments  are  made  in  various  sizes, 
from  the  size  of  a  large  watch  up  to  an  8  or  10  inch  face. 
The  barometer  consists  of  a  cylindrical  box  of  metal,  with  a 
top  of  thin,  elastic,  corrugated  metal.     The  air  is  removed 
from  the  box.     When  the  atmospheric  pressure  increases, 
the  top  is  pressed  inwards,  and  when  it  is  diminished,  the 
top  is  pressed   outwards   by  its  own  elasticity,  aided  by  a 
spring  beneath.     These  movements  of  the  cover  are  trans- 
mitted and  multiplied  by  a  combination  of  delicate  levers 
that  act  upon  an  index  hand  and  cause  it  to  move  either  to 
the  right  or  left  over  a  graduated  scale.     These  barometers 
are  self -correcting  (compensated)  for  variations  in  tempera- 
ture.    They  are  very  portable,  occupying  but  a  small  space, 
and  are  so  delicate  that  they  are  said  to  show  a  difference 
in  the  atmospheric  pressure  when  transferred  from  the  table 
to  the  floor.     They  must  be  handled  with  care,  as  they  are 
easily  injured.     The  mercurial  barometer  is  the  standard. 

53.  With  air,  as  with  water,  the  lower  we  go,  the  greater 
the  pressure,  and  the  higher  we  go,  the  less  the  pressure. 
At  the  level  of  the  sea,  the  height  of  the  mercurial  column 
is  about  30  inches;  at  5,000  feet  above  the  sea,  it  is  24.7 
inches;  at  10,000  feet,  it  is  20.5  inches;  at  15,000  feet,  it  is 
16.9  inches;  at  3  miles,  it  is  16.4  inches;  and  at  6  miles,  it  is 
8.9  inches. 

54.  The  density  also  varies  with  the  altitude;  that  is,  a 
cubic  foot  of  air  at  an  elevation  of  5,000  feet  above  the  sea 
level  will  not  weigh  as  much  as  a  cubic  foot  at  sea  level. 
This  is  proved  conclusively  by  the  fact  that  at  a  height  of  3| 
miles  the  mercurial  column  measures  but  15  inches,  indi- 
cating that  half  the  weight  of  the  entire  atmosphere  is  below 


24  PHYSICS.  §  4 

that.  It  is  known  that  the  height  of  the  earth's  atmosphere 
is  at  least  50  miles;  hence,  the  air  just  before  reaching  the 
limit  must  be  in  an  exceedingly  rarefied  state.  It  is  by 
means  of  barometers  that  great  heights  are  measured.  The 
aneroid  barometer  has  the  heights  marked  on  the  dial,  so 
that  it  can  be  read  directly.  With  the  mercurial  barometer 
the  heights  must  be  calculated  from  the  reading. 

55.  The  atmospheric  pressure  is  everywhere  present,  and 
presses  all  objects  in  all  directions  with  equal  force.  If  a 
book  is  laid  upon  the  table,  the  air  presses  upon  it  in  every 
direction  with  an  equal  average  force  of  14. 7  pounds  per 
square  inch.  It  would  seem  as  though  it  would  take  con- 
siderable force  to  raise  a  book  from  the  table,  since  if  the 
size  of  the  book  were  8  inches  by  5  inches,  the  pressure  upon 
it  is  8  X  5  X 14. 7  =  588  pounds ;  but  there  is  an  equal  pressure 
beneath  the  book  to  counteract  the  pressure  on  the  top.  It 
would  now  seem  as  though  it  would  require  a  great  force  to 
open  the  book,  since  there  are  two  pressures  of  588  pounds 
each,  acting  in  opposite  directions,  and  tending  to  crush  the 
book;  so  it  would,  but  for  the  fact  that  there  is  a  layer  of 
air  between  each  leaf,  acting  upwards  and  downwards  with  a 
pressure  of  14. 7  pounds  per  square  inch. 

If  two  metal  plates  are  made  as  perfectly  smooth  and  flat 
as  it  is  possible  to  get  them,  and  the  edge  of  one  be  laid 
upon  the  edge  of  the  other,  so  that  one  may  be  slid  upon  the 
other  and  the  air  thus  excluded,  it  will  take  an  immense  force, 
compared  with  the  weight  of  the  plates,  to  separate  them. 
This  is  because  the  full  pressure  of  14.7  pounds  per  square 
inch  is  then  exerted  upon  each  plate,  with  no  counteracting 
equal  pressure  between  them. 

If  a  piece  of  flat  glass  is  laid  upon  a  flat  surface  that  has 
been  previously  moistened  with  water,  it  will  require  consid- 
erable force  to  separate  them ;  this  is  because  the  water  helps 
to  fill  up  the  pores  in  the  flat  surface  and  in  the  glass,  and 
thus  creates  a  partial  vacuum  between  the  glass  and  the 
surface,  thereby  reducing  the  counterpressure  beneath  the 
glass. 


§  4  PHYSICS.  25 

EXPANSION  OF  GASES. 

56.  In  Fig.  8  the  space  above  the  column  of  mercury 
was  said  to  be  a  vacuum,  and  that  if  any  gas  or  air  was 
present,  it  would  expand,  its  tension  forcing  the  column  of 
mercury  downwards.  If  enough  gas  is  admitted  to  cause 
the  mercury  to  stand  at  15  inches,  the  tension  of  the  gas  is 

14  7 
evidently          -  =  7.35   pounds   per   square  inch,  since  the 

2 

pressure  of  the  outside  air  of  14.7  pounds  per  square  inch 
only  balances  15  inches,  instead  of  30  inches  of  mercury;  that 
is,  it  balances  only  half  as  much  as  it  would  if  there  were  no 
gas  in  the  tube;  therefore,  the  pressure  (tension)  of  the  gas 
in  the  tube  is  7.35  potinds.  If  more  gas  is  admitted,  until 
the  top  of  the  mercurial  column  is  just  level  with  the  mer- 
cury in  the  cup,  the  gas  in  the  tube  has  then  a  tension  equal 
to  the  outside  pressure  of  the  atmosphere.  Suppose  that 
the  bottom  of^the  tube  is  fitted  with  a  piston,  and  that  the 
total  length  of  the  inside  of  the  tube  is  36  inches.  If  the 
piston  is  shoved  upwards,  so  that  the  space  occupied  by  the 
gas  is  18  inches  long,  instead  of  36  inches,  the  temperature 
remaining  the  same  as  before,  it  will  be  found  that  the  ten- 
sion of  the  gas  within  the  tube  is  29.4  pounds  per  square 
inch.  It  will  be  noticed  that  the  volume  occupied  by  the 
gas  is  only  half  that  in  the  tube  before  the  piston  was  moved, 
while  the  pressure  is  twice  as  great,  since  14.7x2  =  29.4 
pounds.  If  the  piston  is  pushed  farther  up,  so  that  the  space 
occupied  by  the  gas  is  only  9  inches,  instead  of  18  inches,  the 
temperature  still  remaining  the  same,  the  pressure  will  be 
found  to  be  58.8  pounds  per  square  inch.  The  volume  has 
again  been  reduced  one-half,  and  the  pressure  increased  2 
times,  since  29.4x2  =  58.8  pounds.  The  space  now  occu- 
pied by  the  gas  is  9  inches  long,  whereas,  before  the  piston 
was  moved,  it  was  36  inches  long;  as  the  tube  was  assumed 
to  be  of  uniform  diameter  throughout  its  length,  the  volume 
is  now  -^g-  =  \  of  its  original  volume,  and  its  pressure  is 

58  8 

_l_  =  4  times  its  original  pressure.      Moreover,  if  the  tem- 
perature of  the  confined  gas  remains  the  same,  the  pressure 


26  PHYvSICS.  §  4 

and  volume  will  always  vary  in  a  similar  way.    The  law  that 
states  these  effects  is  called  Mariotte's  /aw,  and  is  as  follows : 

57.  Marietta's  Law.—  The  temperature  remaining  the 
same,  tJie  volume  of  a  given  quantity  of  gas  varies  inversely 
as  the  pressure. 

The  meaning  of  this  is:  If  the  volume  of  the  gas  is  dimin- 
ished to  i,  i,  |,  etc.  of  its  former  volume,  the  tension  will 
be  increased  2,  3,  5,  etc.  times ;  or,  if  the  outside  pressure  is 
increased  2,  3,  5,  etc.  times,  the  volume  of  the  gas  will  be 
diminished  to  ^,  ^,  J,  etc.  of  its  original  volume,  the  tem- 
perature remaining  constant.  It  also  means  that,  if  a  gas  is 
under  a  certain  pressure,  and  the  pressure  is  diminished  to 
i,  i,  yL-,  etc.  of  its  original  pressure,  the  volume  of  the  con- 
fined gas  will  be  increased  2,  3,  10,  etc.  times — its  tension 
decreasing  at  the  same  rate. 

Suppose  3  cubic  feet  of  air  to  be  under  a  pressure  of 
60  pounds  per  square  inch  in  a  cylinder  fitted  with  a  mov- 
able piston ;  then,  the  product  of  the  volume  and  pressure  is 
3  X  60  =  180.  Let  the  volume  be  increased  to  6  cubic  feet, 
then  the  pressure  will  be  30  pounds  per  square  inch,  and 
30x6  =  180,  as  before.  Let  the  volume  be  increased  to  24 
cubic  feet,  it  is  then  -2^-  —  8  times  its  original  volume,  and  the 
pressure  is  \  of  its  original  pressure,  or  60  X  \  =  7^-  pounds, 
and24x7|-  =  180,  as  in  the  two  preceding  cases.  It  will 
now  be  noticed  that  if  a  gas  is  enclosed  within  a  confined 
space,  and  allowed  to  expand  without  losing  any  heat,  the 
product  of  the  pressure  and  the  corresponding  volume  for  one 
position  of  tlie  piston  is  the  same  as  for  any  other  position  of 
the  piston.  If  the  piston  were  to  compress  the  air,  the  same 
result  would  be  obtained. 

Let     p    =  pressiire  for  one  position  of  the  piston ; 

pl  =  pressure  for  any  other  position  of  the  piston  ; 
v    =  volume  corresponding  to  the  pressure/; 
7'j  =  volume  corresponding  to  the  pressure  p^. 

Then,  /*>  =/,«',-  (?•) 

Knowing  the  volume  and  the  pressure  for  any  position  of 
the  piston  and  the  volume  for  any  other  position,  the  pressure 


§  4  PHYSICS.  27 

may  be  calculated,  or  if  the  pressure  is  known  for  any  other 
position,  the  volume  may  be  calculated. 

EXAMPLE  i.  —  If  1.875  cubic  feet  of  air  be  under  a  pressure  of  12  pounds 
per  square  inch,  what  will  be  the  pressure  when  the  volume  is  increased 

(a)  to  2  cubic  feet  ?  (b]  to  3  cubic  feet  ?  (c]  to  9  cubic  feet  ? 

SOLUTION.  —  Solving  formula  7  for/i,  the  unknown  pressure, 

pv       72x1-875 

(a)  pl  —  *-—  —  -  ^-=  --  =  67|  Ib.  per  sq.  in.     Ans. 

79  V  1  87^ 

(b)  p,  =  -  ^^  -  =  45  Ib.  per  sq.  in.     Ans. 

(c)  pi  =  72x*'875  =  15  ib.  per  sq.  in.     Ans. 

EXAMPLE  2.  —     10  cubic  feet  of  air  have  a  tension  of  5.6  pounds  per 
square  inch  ;  what  is  the  volume  when  the  tension  is  (a)  4  pounds  ? 

(b)  8  pounds  ?  (c)  25  pounds  ?  (d)  100  pounds  ? 

SOLUTION.  —  Solving  formula  7  for  vlt 

(a)  y.=          =  5-6*10  =  14cu.ft 


(b)  Vi  =  5-6*10  =  7  cu.  ft.     Ans. 

(c)  Vi  =  5'6  *  10  =  2.24cu.  ft.     Ans. 

25 

(d)  vt  =  5'61^)1°  =".66  cu.  ft.     Ans. 

58.  There  are  two  ways  of  measuring  the  pressure  of  a 
gas:  by  means  of  an  instrument  called  a  manometer,  and  by 
means  of  a  gauge.  The  manometer  generally  used  is  practi- 
cally the  same  as  a  mercurial  barometer,  except  that  the  tube 
is  much  longer,  so  that  pressures  equal  to  several  atmos- 
pheres may  be  measured,  and  is  enlarged  and  bent  into  a 
U  shape  at  the  lower  end  ;  both  lower  and  upper  ends  are 
open,  the  lower  end  being  connected  to  the  vessel  containing 
the  gas,  whose  pressure  it  is  desired  to  measure.  The  gauge 
is  so  common  that  no  description  of  it  will  be  given  here. 
With  both  the  manometer  described  above  and  the  gauge, 
the  pressures  recorded  are  the  amounts  by  which  they  exceed 
the  atmospheric  pressure,  and  are  called  the  gauge  pressures. 
To  find  the  real  pressure,  called  the  absolute  pressure,  the 


28  PHYSICS.  §  4 

atmospheric  pressure  must  be  added  to  the  gauge  pressure. 
In  all  formulas  in  which  the  pressure  of  a  gas  or  steam  is 
used,  the  absolute  pressure  must  be  used,  unless  the  gauge 
pressure  is  distinctly  specified  as  being  the  proper  pressure 
to  use.  For  convenience,  all  pressures  given  in  this  Instruc- 
tion Paper  and  in  the  accompanying  Question  Paper  will  be 
absolute  pressures,  and  the  word  absolute  will  be  omitted  to 
avoid  its  constant  repetition. 

59.  As  a  necessary  consequence   of  Mariotte's  law,   it 
may  be  stated  that  the  density  of  a  gas  varies  directly  as  the 
pressure,  and  inversely  as  the  volume  ;   that  is,  the  density 
increases  as  the  pressure  increases,  and  decreases  as  the  vol- 
ume increases 

This  is  evident,  since,  if  a  gas  has  a  tension  of  2  atmos- 
pheres, or  14.7x2  =  29.4  pounds  per  square  inch,  it  will 
weigh  twice  as  much  as  the  same  volume  would  if  the  ten- 
sion were  1  atmosphere,  or  14.7  pounds  per  square  inch. 
For,  let  the  volume  be  increased  until  it  is  twice  as  great  as 
the  original  volume,  the  tension  will  then  be  1  atmosphere. 
The  total  weight  of  the  gas  has  not  been  changed,  but  there 
are  now  2  cubic  feet  for  every  cubic  foot  of  the  original  vol- 
ume, and  the  weight  of  1  cubic  foot  now  is  only  half  as 
great  as  before.  Thus,  the  density  decreases  as  the  volume 
increases,  and  as  an  increase  of  pressure  causes  a  decrease  of 
volume,  the  density  increases  as  the  pressure  increases. 

Let     D    —  density  corresponding  to  the  pressure  /  and 

volume  v; 

D1  —  density  corresponding  to  the  pressure  /1  and 
volume  z>r 

Then,  /  :  D   =  pl :  D^  or/£>,  •=  /,  D,  (8.) 

and  v:D^  =  v^.D,  or  v  D   =  *;,/?,.  (9.) 

60.  Since  the  weight  is  proportional  to  the  density,  the 
weights  may  be  used  in  place  of  the  densities  in  formulas  8 
and  9. 


§  4  PHYSICS.  29 

Let     W  =  weight  of  a  cubic  foot  of  air  or  other  gas, 

whose  volume  is  v  and  pressure  is  /  ; 
W^  =  weight  of  a  cubic  foot  when  the  volume  is  vl 
and  pressure  is^. 

Then,  pWl=plW.      .      (1C.) 

vW  =  v^Wv.  (11.) 

EXAMPLE  1.  —  The  weight  of  1  cubic  foot  of  air  at  a  temperature  of 
60°  F.,  and  under  a  pressure  of  1  atmosphere  (14.7  pounds  per  square 
inch),  is  .0763  pound;  what  will  be  the  weight  per  cubic  foot  if  the  air 
is  compressed  until  the  tension  is  5  atmospheres,  the  temperature  still 
being  60°  F.  ? 

SOLUTION.  —  Applying  formula  1O,  p  Wi  =  pi  W, 
or  IX^i  =  5  X-  0763. 

Hence,  Wl  =  .3815  Ib.  per  cu.  ft.     Ans. 

EXAMPLE  2.  —  If  in  the  last  example  the  air  had  expanded  until  the 
tension  was  5  pounds  per  square  inch,  what  would  have  been  its  weight 
per  cubic  foot  ? 

SOLUTION.  —  Applying  formula  1O,  p  Wi  —  pi  W.  Here,/  =  14.7, 
/,  =  5,  and  W  =  .0763.  Hence,  14.7  X  W^  =  5  X  .0763, 

=  .02595  Ib.  per  cu.  ft.     Ans. 


EXAMPLE  3.  —  If  6.75  cubic  feet  of  air  at  a  temperature  of  60°  F.,  and 
a  pressure  of  1  atmosphere,  are  compressed  to  2.25  cubic  feet  (the  tem- 
perature still  remaining  60°  F.),  what  is  the  weight  of  a  cubic  foot  of 
the  compressed  air  ? 

SOLUTION.'  —  Applying  formula  11, 

•v  W  —  V!  Wlt  or  6.75  X  .0763  =  2.25  X  Wi. 

Hence,          W,  =  6'750X'°768  =  .2289  Ib.  per  cu.  ft.     Ans. 


61.  In  all  that  has  been  said  before,  it  has  been  stated 
that  the  temperature  was  constant;  the  reason  for  this  will 
now  be  explained.  Suppose  5  cubic  feet  of  air  to  be  confined 
in  a  cylinder  placed  in  a  vacuum,  so  that  there  will  be  no 
pressure  due  to  the  atmosphere,  and  suppose  the  cylinder  to 
be  fitted  with  a  piston  weighing,  say,  100  pounds  and  having 
an  area  of  10  square  inches.  The  tension  of  the  gas  will  be 
iff.  —  10  pounds  per  square  inch.  Suppose  that  the  tem- 
perature of  the  air  is  32°  F.,  and  that  it  is  heated  until  the 


30  PHYSICS.  §  4 

temperature  is  33°  F.,  i.  e.,  until  the  temperature  is  increased 
1°,  it  will  be  found  that  the  piston  has  risen  a  certain  amount, 
and,  consequently,  the  volume  has  increased,  while  the  pres- 
sure is  the  same  as  before,  or  10  pounds  per  square  inch. 
If  more  heat  is  applied,  until  the  temperature  of  the  gas 
is  34°  F.,  it  will  be  found  that  the  piston  has  again  risen, 
and  the  volume  again  increased,  while  the  pressure  still 
remains  the  same.  It  will  be  found  that  for  every  increase 
of  temperature  there  will  be  a  corresponding  increase  of 
volume.  The  law  that  expresses  this  change  is  called  Gay- 
Lussac's  /aw,  and  is  as  follows: 

62.  Gay-Lussac's  Law.  —  If  the  pressure  remains  con- 
stant, every  increase  of  temperature  of  1°  F.  produces  in  a 
given  quantity  of  gas  an  expansion  of  -^-^  of  its  volume  at 
32°  F. 

If   the   pressure   remains  constant,  it  will  also  be  found 
that   every  decrease  of  temperature  of  1°  F.   will  cause  a 
decrease  of  ^-^  °f  the  volume  at  32°  F. 
Let   v    =  original  volume  of  gas; 
vt  —  final  volume  of  gas  ; 
t    =f  temperature  corresponding  to  volume  v\ 
tv   =  temperature  corresponding  to  volume  v^ 

<»•> 

That  is,  the  volume  of  gas  after  heating  (or  cooling]  is  equal 
to  the  original  volume  multiplied  by  460  plus  the  final  temper- 
ature, divided  by  1+60  plus  the  original  temperature. 

.EXAMPLE.  —    5  cubic  feet  of  air  at  a  temperature  of  45°  are  heated 
under  constant  pressure  up  to  177°  ;  what  is  its  final  volume  ? 
SOLUTION.  —  Applying  formula  13, 

\        K  /460  +  177\ 

)  =  5  (  460T45  )  =  6'307  CU"  ft'     AnS' 


63.  Suppose  a  certain  volume  of  gas  to  be  confined  in  a 
vessel  so  that  it  cannot  expand  ;  in  other  words,  suppose  the 
piston  of  the  cylinder  before  mentioned  to  be  so  fastened 
that  it  cannot  move.  Let  a  gauge  be  placed  on  the  cylinder 
so  that  the  tension  of  the  confined  gas  can  be  registered.  If 


§  4  PHYSICS.  31 

the  gas  is  heated,  it  will  be  found  that  for  every  increase  of 
temperature  of  1°  F.  there  will  be  a  corresponding  increase 
of  ^3-  of  the  tension.  That  is,  the  volume  remaining  con- 
stant, the  tension  increases  ^^  °f  tne  original  tension  for 
every  degree  rise  of  temperature. 

Let  p    =  original  tension; 

t    =  corresponding  temperature; 

/j  =  final  tension; 

/,   =  final  temperature. 


That  is,  if  a  certain  quantity  of  gas  be  heated  (or  cooled) 
from  t°  to  /,°,  the  volume  remaining  constant,  the  resulting 
tension  pl  will  be  equal  to  the  original  tension  multiplied  by 
JfiO  plus  the  final  temperature,  divided  by  4.60  plus  the  original 
temperature. 

EXAMPLE.  —  If  a  certain  quantity  of  air  is  heated  under  constant  vol- 
ume from  45°  to  177°,  what  is  the  resulting  tension,  the  original  tension 
being  14.7  pounds  per  square  inch  ? 

SOLUTION.  —  Applying  formula  13, 


460T45 

64.  Absolute  Temperature.  —  According  to  the  modern 
and  now  generally  accepted  theory  of  heat,  the  atoms  and 
molecules  of  all  bodies  are  in  an  incessant  state  of  vibration. 
The  vibratory  movement  in  the  liquids  is  faster  than  in  the 
solids,  and  in  the  gases,  faster  than  in  either  of  the  other  two. 
Any  increase  of  heat  increases  the  vibrations,  and  a  decrease 
of  heat  decreases  them.  From  experiments  and  calculations, 
it  has  been  concluded  that  at  460°  below  zero  on  the  Fahren- 
heit scale  all  these  vibrations  cease.  This  temperature  is 
called  the  absolute  zero,  and  all  temperatures  reckoned  from 
this  point  are  called  the  absolute  temperatures.  The  point 
of  absolute  zero  has  never  been  reached,  the  lowest  recorded 
temperature  being  about  393°  F.  below  zero,  but,  neverthe- 
less, it  has  a  meaning,  and  is  used  in  many  formulas,  being 
nearly  always  denoted  by  T.  The  ordinary  temperatures 


32  PHYSICS.  §  4 

are  denoted  by  /.  When  the  word  temperature  alone  is 
used,  the  meaning"  is  the  same  as  ordinarily  used,  but  when 
absolute  temperature  is  specified,  460°  F.  must  be  added  to 
the  ordinary  temperature.  The  absolute  temperature  cor- 
responding to  212°  F.  is  460° +  212°  =  072°  F.  If  the  abso- 
lute temperature  is  given,  the  ordinary  temperature  may  be 
found  by  subtracting  460°  from  the  absolute  temperature. 
For  example,  the  absolute  temperature  being  520°  F.,  the 
ordinary  temperature  is  520°  -  460°  =  60°. 

65.  The   relation  between   the   pressure,   volume,    and 
temperature  of  air  is  given  by  the  following  formula: 

Let  p  =  pressure  of  air  in  pounds  per  square  inch ; 
v  =  volume  of  1  pound  of  air  in  cubic  feet; 
V  =  volume  of  W  pounds  of  air  in  cubic  feet ; 
T  —  absolute  temperature; 
W  —  weight  of  air  in  pounds. 
Then,  pv  —  .37052  T.  (14.) 

That  is,  the  pressure  in  pounds  per  square  inch,  multiplied 
by  the  volume  of  a  pound  of  air  in  cubic  feet,  is  equal  to 
.37052  times  the  absolute  temperature  corresponding  to  the 
pressure  p  and  volume  v. 

In  this  formula  the  weight  of  the  air  is  1  pound. 

EXAMPLE  1. — The  pressure  upon  9  cubic  feet  of  air  weighing  1  pound 
is  20  pounds  per  square  inch ;  what  is  the  temperature  ? 
SOLUTION. — Applying  formula  14, 

p  v  =  .37052  T,  or  20  X  9  =  .37052  T. 

180 
Hence,  T  =    0    "     =  485.8°,  nearly. 

.  o  i  UOrO 

485.8°  —  460°  =  25.8°,  the  temperature.     Ans. 

EXAMPLE  2. — What  is  the  volume  of  1  pound  of  air  whose  tempera- 
ture is  60°  F.  under  a  pressure  of  1  atmosphere  ? 

SOLUTION. — Applying  formula  14, p  v  =  .37052  T,  and  substituting, 
14.7  X  v  =  .37052  X  (460  +  60)  =  .37052  X  520, 

.37052X520  ,.        A 

or  it  = ^-~ =  13.107cu.  ft.     Ans. 

66.  If  the  weight   of  the   air  is   greater   or   less  than 
1  pound,  the  following  formula  must  be  used: 

pV  =  .37052  WT.  (15.) 


§  4  PHYSICS.  33 

That  is,  tJic  pressure  in  pounds  per  square  inch  multiplied 
by  tJie  volume  in  eubic  feet  is  equal  to  .3705%  times  the  weight 
in  pounds  multiplied  by  the  absolute  temperature. 

EXAMPLE  1.  —  3  cubic  feet  of  air  weighing  .35  pound  are  under  a 
pressure  of  48  pounds  per  square  inch  ;  what  is  the  temperature  of  the 
air  ? 

SOLUTION.  —  Applying   formula    15,  /  V  —  .37052  W  T,  and   substi- 

48X3  =  .  37052  X.  35  X  T, 

-  =  '•"o-4- 

Then,  1,110.4°  -460°  =  650.4°.     Ans. 

EXAMPLE  2.  —  What  is  the  weight  of  1  cubic  foot  of  air  at  a  tempera- 
ture of  32°  ,  and  under  a  pressure  of  1  atmosphere  ? 

SOLUTION.  —  Applying   formula  15,  p  V  •=  .37052  W7  7",  and   substi- 
14.7  X  1  =  -37052  X  (460  +  32)  X  W, 


If  .the  pressure  be  taken  as  14.69856  pounds  per  square  inch,  and  the 
absolute  zero  as  459.4°  instead  of  460°  below  zero,  and  if  .370514  be  used 
instead  of  .37052  (more  exact  values),  the  weight  of  1  cubic  foot  is 

=.  08073  ,„..  nearly. 

EXAMPLE  3.  —  What  is  the  exact  volume  of  1  pound  of  air  at  a  tem- 
perature of  32°,  and  at  a  pressure  of  1  atmosphere?  Take  absolute 
zero  as  459.4°,  and  the  pressure  as  14.69856  pounds  per  square  inch. 

SOLUTION.—  p  V  =  .370514  W  T, 

or  14.69856  X  V  =  .370514  X  1  X  (459.4  +  32). 


67.     If  in  the  formula  p  V  —  .  37052  W  T,  both  sides  of  the 
equation  be  divided  by  T  (which,  of  course,  does  not  alter 

pV 
the  equality),  there  results  the  expression  *-=-  =  .  37052  W. 

Let  /,,  Fj,  and  71,  represent  the  pressure,  volume,  and 
temperature  of  the  same  weight  of  air  in  another  state; 
then,  /,Ft  =  .37052  WTX.  Dividing  both  sides  by  71,, 

-  =  .37052  W.     Therefore,  since          and  are  equal 


34  PHYSICS.  §  4 

to  the  same  thing  (i.  e.,  .  37052  W),  they  are  equal  to  each 
other,  and  ,  y        py 

£f=£f±.  (16.) 

This  very  important  formula  is  the  complete  expression  of 
Gay-Lussac's  law,  and  is  true  for  any  of  .the  so  called  perma- 
nent gases.  It  was  from  this  formula  that  formulas  13  and 
13.  were  derived.  Thus,  let  the  pressure  be  constant;  then, 


larly,  letting  the  volume  be  constant,  V  •=  F,,  and  ^~-  =  OFT, 
or  A  =  ^  =  P  (^^7)-      So>  also>  by  letting  the  tem- 

perature be  constant,  T  —  7\  and  ^~-  =     *    ',  or  /  V  —  /,  F,, 

which  is  the  same  as  formula  7. 

In  formulas  7,  16,  17,  and  18,  it  matters  not  with  what 
units  the  pressures  and  volumes  are  measured,  except  that 
they  must  be  the  same  throughout  the  same  example,  and 
the  pressures  must  always  be  absolute  pressures. 


EXAMPLES  FOR  PRACTICE. 

68.      Solve  the  following: 

1.  A  vessel  contains  25  cubic  feet  of  gas  at  a  pressure  of  18  pounds 
per  square  inch ;  if  125  cubic  feet  of  gas  having  the  same  pressure  are 
forced  into  the  vessel,  what  will  be  the  resulting  pressure  ? 

Ans.   108  Ib.  per  sq.  in. 

2.  A  pound  of  air  has  a  temperature  of  126°  and  a  pressure  of 
1  atmosphere;  what  volume  does  it  occupy  ?  Ans.   14.77  cu.  ft. 

.  3.  A  certain  quantity  of  air  has  a  volume  of  26.7  cubic  feet,  a  pres- 
sure of  19.3  pounds  per  square  inch,  and  a  temperature  of  42° ;  what  is 
the  weight  ?  Ans.  2.77  Ib. 

4.  A   receiver   contains   180  cubic    feet    of  gas   at   a  pressure  of 
20  pounds  per  square  inch ;  if  a  vessel  holding  12  cubic  feet  be  filled 
from  the  receiver  until  its  pressure  is  20  pounds  per  square  inch,  what 
will  be  the  pressure  in  the  receiver  ?  Ans.  18§  Ib.  per  sq.  in. 

5.  If  10  cubic  feet  of  air  having  a  pressure  of  22  pounds  per  square 
inch  and  a  temperature  of  75°  are  heated  until  the  temperature  is  300°, 
the  volume  remaining  the  same,  what  is  the  new  pressure  ? 

Ans.  31.251b.  per  sq.  in. 


§4  PHYSICS.  35 

MIXTURES  OF  GASES. 

69.  If  two  liquids  that  do  not  act  chemically  on  each 
other  are   mixed  together  and  allowed  to  stand,  it  will  be 
found  that  after  a  time  the  two  liquids  have  separated,  and 
that  the  heavier  has  fallen  to  the  bottom.     If  two  vessels  of 
equal  capacities  containing  gases    of  different  densities  are 
put  in  communication  with  each  other,  the  gases  will  be  found 
to  have  mixed  in  equal  proportions  after  a  short  time.     If 
one  vessel  is  higher  than  the  other,  and  the  heavier  gas  is  in 
the  lower  vessel,  the  same  result  will  occur.     The  greater 
the  difference  of  the  densities  of  the  two  gases,  the  quicker 
they  will  mix.     It  is  assumed  that  no  chemical  action  takes 
place  between  the  two  gases.     When  the  two  gases  have  the 
same  temperature  and  pressure,  the  pressure  of  the  mixture 
will  be  the  same ;  this  is  evident,  since  the  total  volume  has 
not  been  changed,  and  unless  the  volume  or  temperature 
changes,  the  pressure  cannot  change.     This  property  of  the 
mixing  of  gases  is  a  very  valuable  one,  since,  if  they  acted 
like  liquids,   carbonic-acid  gas  (the  result  of   combustion), 
which  is  1^  times  as  heavy  as  air,  would  remain  next  to  the 
earth  instead  of  dispersing  into  the  atmosphere,  the  result 
being  that  no  animal  life  could  exist. 

70.  Mixture    of  Equal  Volumes  of  Gases   Having 
Unequal  Pressures. — If  two  gases  Jiaving  equal  volumes 
and  temperatures,   but    different  pressures,  are  mixed  in  a 
vessel  whose  volume  is  equal  to  one  of  the  equal  volumes  of 
the  gas,  the  pressure  of  the  mixture  will  be  equal  to  the  sum 
of  the  two  pressures,  provided  that  the  temperature  remains 
the  same  as  before. 

EXAMPLE. — Two  vessels  containing  3  cubic  feet  of  gas,  each  at 
a  temperature  of  60°  and  subjected  to  pressures  of  40  pounds  and 
25  pounds  per  square  inch,  respectively,  are  placed  in  communication 
with  each  other,  and  all  the  gas  is  compressed  into  one  vessel.  If  the 
temperature  of  the  mixture  is  also  60°,  what  is  the  pressure  ? 

SOLUTION. — According  to  the  rule  just  given,  the  pressure  will  be 
40  +  25  =  65  pounds  per  square  inch.  This  may  be  proved  by  appli- 
cations of  Mariotte's  law  ;  thus,  compress  the  gas  whose  pressure  is 
25  pounds  per  square  inch,  until  its  pressure  is  40  pounds ;  its  volume 


36  PHYvSICS.  §4 

may  be  found  thus:  p  v  =  p\Vi,  or  25  X  3  =  40  X  ^  ;  whence,  v  =  1.875 
cubic  feet.  Let  communication  be  established  between  the  two  ves- 
sels. The  pressure  will  evidently  be  40  pounds,  and  the  total  volume 
3  +  1.875  =  4.875  cubic  feet.  If  this  be  compressed  until  the  volume  is 
3  cubic  feet,  the  temperature  remaining  at  60°  throughout  the  whole 
operation,  the  final  pressure  may  be  found  by  formula  7,  p  v  —  p\  VL 

Thus,  40  X  4.875  =  pl  X  3,  and/!  =  -  X4.875  =  65  pounds  per  square 
inch,  as  before. 

71.  Mixture  of  Two  Gases  HaAdng  Unequal  Vol- 
umes and  Pressures. 

Let  v  and  p    =  volume  and  pressure,  respectively,  of  one 

of  the  gases; 
vl  and  pl  =  volume  and  pressure,  respectively,  of  the 

other  gas; 
Fand  P  =  volume  and  pressure,  respectively,  of  the 

mixture. 
Then,  if  the  temperature  remains  the  same, 

VP=vp  +  vJr  (17.) 

That  is,  if  the  temperature  is  constant,  the  volume  after 
mixture,  multiplied  by  the  resulting  pressure,  is  equal  to  the 
volume  of  one  gas  before  mixture  multiplied  by  its  pressure, 
plus  the  volume  of  the  other  gas  multiplied  by  its  pressure. 

EXAMPLE.  —  Two  gases  of  the  same  temperature,  having  volumes  of 
7  cubic  feet  and  4|  cubic  feet,  and  whose  pressures  are  27  pounds  and 
18  pounds  per  square  inch,  respectively,  are  mixed  together  in  a  vessel 
whose  volume  is  10  cubic  feet.  The  temperature  of  the  two  gases  and 
of  the  mixture  being  60°  F.  ,  what  is  the  resulting  pressure  ? 

SOLUTION.  —  Applying  formula  17, 
PV  = 


Hence,  P  =  =  27  Ib.  per  sq.  in.     Ans. 

7  2  .  Mixtu  re  of  Two  Vol  times  of  Air  Having  Unequal 
Pressures,  Volumes,  and  Temperatures.  —  If  a  body  of  air 
having  a  temperature  flt  a  pressure  /,,  and  a  volume  v^  is 
mixed  with  another  volume  of  air  having  a  temperature 
/2,  a  pressure  /a,  and  a  volume  z>2,  and  the  mixture  has  a 
volume  V,  pressure  P,  and  a  temperature  t,  then,  either  the 
new  temperature  /,  the  new  volume  V,  or  the  new  pressure  P 
may  be  found,  if  the  other  two  quantities  are  known,  by  the 


§  4  PHYSICS.  37 

following1  formula,  in  which  T^  T^  and  T  are  the  absolute 
temperatures  corresponding  to  /J5  /2,  and  /: 


EXAMPLE.  —  5  cubic  feet  of  air  having  a  tension  of  30  pounds  per 
square  inch,  and  a  temperature  of  80°  F.  ,  are  required  to  be  compressed 
together  with  11  cubic  feet  of  air  having  a  tension  of  21  pounds  per 
square  inch,  and  a  temperature  of  45°  F.,  in  a  vessel  whose  cubical 
contents  are  8  cubic  feet.  The  new  pressure  is  required  to  be  45  pounds 
per  square  inch.  What  is  the  temperature  of  the  mixture  ? 

SOLUTION.  —  Substituting  in  formula  18, 


Hence,         T  =  -_  =  489.66°,  nearly,  and  /  =  29.66°.     Ans. 


EXAMPLES    FOR   PRACTICE. 

73.      Solve  the  following: 

1.  Two  vessels  contain  air  at  pressures  of  60  and  83  pounds  per 
square  inch,  respectively.    The  volume  of  each  vessel  is  8.47  cubic  feet. 
If  all  of  the  air  in  both  vessels  is  removed  to  another  vessel,  and  the 
new  pressure  is  100  pounds  per  square  inch,  what  is  the  volume  of  the 
vessel,  the  temperature  being  the  same  throughout  ?    Ans.   12.11  cu.  ft. 

2.  A  vessel  contains  11.83  cubic  feet  of  air  at  a  pressure  of  33.3 
pounds  per  square  inch.     It  is  desired  to  increase  the  pressure  to  40 
pounds  per  square  inch  by  supplying  air  from  a  second  vessel  that 
contains  19.6  cubic  feet  of  air  at  a  pressure  of  60  pounds  per  square 
inch.    What  will  be  the  pressure  in  the  second  vessel  after  the  pressure 
in  the  first  has  been  raised  to  40  pounds  per  square  inch  ? 

Ans.  55.96  Ib.  per  sq.  in. 

3.  If  4.8  cubic  feet  of  air  having  a  tension  of  52  pounds  per  square 
inch  and  a  temperature  of  170°  are  mixed  with  13  cubic  feet  of  air  hav- 
ing a  tension  of  78  pounds  per  square  inch  and  a  temperature  of  265°, 
what  must  be  the  volume  of  the  vessel  containing  the  mixture,  in  order 
that  the  tension  of  the  mixture  may  be  30  pounds  per  square  inch  and 
the  temperature  80°  ?  Ans.  32.31  cu.  ft. 


THE    AIR   PUMP. 

74.  The  air  pump  is  an  instrument  for  removing  air 
from  an  enclosed  space.  A  section  of  the  principal  parts  is 
shown  in  Fig.  11,  and  the  complete  instrument  in  Fig.  12. 


38 


PHYSICS. 


The  closed  vessel  R  is  called  the  receiver,  and  the   space 
enclosed  by  it  is  that  from  which  it  is  desired  to  remove  the 


air. 


FIG.  11. 

The  receiver  is  usually  made  of  glass,  and  the  edges 

are  ground  so  as  to 
be  perfectly  air-tight. 
When  made  in  the  form 
shown,  it  is  called  a 
bell-jar  receiver.  The 
receiver  rests  upon  a 
horizontal  plate,  in  the 
center  of  which  is  an 
opening  communica- 
ting with  the  pump 
cylinder  C,  by  means 
of  a  bent  tube  /.  The 
pump  piston  fits  the 
cylinder  accurately,  and 
has  a  valve  V  opening 
upwards.  At  the  junc- 
tion of  the  tube  with 
the  cylinder  is  another 
valve  V  also  opening 
upwards.  When  the 


FIG.  12. 


§  4  PHYSICS.  39 

piston  is  raised  the  valve  V  closes,  and,  since  no  air  can  get 
into  the  cylinder  from  above,  the  piston  leaves  a  vacuum 
behind  it.  The  pressure  on  top  of  V  being  now  removed, 
the  tension  of  the  air  in  the  receiver  R  causes  V  to  rise ;  the 
air  in  the  receiver  then  expands  and  occupies  the  space  dis- 
placed by  the  piston,  the  space  in  the  tube  /  and  in  the 
receiver  R.  The  piston  is  now  pushed  down,  the  valve  V 
closes,  the  valve  V  opens,  and  the  air  in  C  escapes.  The 
lower  valve  Fis  sometimes  supported,  as  shown  in  Fig.  11, 
by  a  metal  rod  passing  through  the  piston  and  fitting  it 
somewhat  tightly.  When  the  piston  is  raised  or  lowered, 
this  rod  moves  with  it.  A  button  near  the  upper  end  of  the 
rod  confines  its  motion  to  within  very  narrow  limits,  the  piston 
sliding  upon  the  rod  during  the  greater  part  of  the  journey. 

75.  Degrees    and   limits    of  Exhaustion. — Suppose 
that  the  volume  of  R  and  /  together  is  four  times  that  of 
C,  and  that  there  are,  say,  200  grains  of  air  in  R  and  t,  and 
50  grains  in  C,  when  the  piston  is  at  the  top  of  the  cylinder. 
At  the  end  of  the  first  stroke,  when  the  piston  is  again  at 
the  top,  50  grains  of  air  in  the  cylinder  C  will  have  been 
removed,  and  the  200  grains  in  R  and  t  will  occupy  the 
spaces  R,  t,  and  C.    The  ratio  between  the  sum  of  the  spaces 
R  and  /,  and  the  total   space  R  +  tC  is  f;   hence,   200  Xf 
=  160  grains  =  the  weight  of  air  in  R  and  /  after  the  first 
stroke.     After  the  second  stroke,  the  weight  of  the  air  in  R 
and  /  would   be   200  X  f  X  |  =  200  X  (4 )2  =  200  X  if  =  128 
grains.     At  the  end  of  the  third  stroke  the  weight  would  be 
[200  X  (|)2]  X  f-  =  200  X  (i)3  =  200  X  Jfo  =  102. 4  grains.     At 
the  end  of  n  strokes  the  weight  would  be  200  X  (I)".     It  is 
evident  that  it  is  impossible  to  remove  all  of  the  air  that  is 
contained  in  R  and  t  by  this  method.     It  requires  an  exceed- 
ingly good  air  pump  to  reduce  the  tension  of  the  air  in  R  to 
^5-  of  an  inch  of  mercury.     When  the  air  has  become  rarefied 
to  this  extent,  the  valve  V  will  not  lift,  and,  consequently, 
no  more  air  can  be  exhausted. 

76.  Sprengel's  Air  Pump. — In  Fig.  13,  c  d  is  a  glass 
tube  longer  than  30  inches,  open  at  both  ends,  and  connected, 


40 


PHYSICS. 


by  means  of  india-rubber  tubing,  with  a  funnel  A  filled  with 
mercury  and  supported  by  a  stand.  Mercury  is  allowed  to 

fall  into  this  tube  at  a  rate 
regulated  by  a  clamp  at  c. 
The  lower  end  of  the  tube 
c  d  fits  in  the  flask  B,  which 
has  a  spout  at  the  side  a 
little  higher  than  the  lower 
end  of  c  d  ;  the  upper  part 
has  a  branch  at  x  to  which 
a  receiver  R  can  be  tightly 
fixed.  When  the  clamp  at 
c  is  opened,  the  first  por- 
tions of  the  mercury  that 
run  out  close  the  tube  and 
prevent  air  from  entering 
from  below.  These  drops 
of  mercury  act  like  little 
pistons,  carrying  the  air  in 
front  of  them  and  forcing 
it  out  through  the  bottom 
of  the  tube.  The  air  in  R 
expands  to  fill  the  tube 
every  time  that  a  drop  of 
mercury  falls,  thus  crea- 
ting a  partial  vacuum  at  R, 
which  becomes  more  nearly 
complete  as  the  process 
goes  on.  The  escaping 
mercury  falls  into  the  dish 
//,  from  which  it  can  be 
FIG-  13-  poured  back  into  the  funnel 

from  time  to  time.  As  the  exhaustion  from  R  goes  on,  the 
mercury  rises  in  the  tube  c  d  until,  when  the  exhaustion  is 
complete,  it  forms  a  continuous  column  30  inches  high;  in 
other  words,  it  is  a  barometer  whose  Torricellian  vacuum  is 


NOTE. — A  theoretically  perfect  vacuum  is  sometimes  called  a  Torri- 
cellian vacuum. 


§  4  PHYSICvS.  41 

the  receiver  R.  This  instrument  necessarily  requires  a  great 
deal  of  time  for  its  operation,  but  the  results  are  very  com- 
plete, a  vacuum  of  ^--J-o-g-  of  an  inch  of  mercury  being 
sometimes  obtained.  By  the  use  of  chemicals  in  addition  to 
the  above,  a  vacuum  of  -g^-oVon  °f  an  inc^  °f  mercury  has 
been  obtained. 


HEAT. 

NATURE    OF    HEAT. 

77.  As  to  the  exact  nature  of  heat,  scientists  differ,  but  all 
modern  thinkers  and  investigators  agree  that  heat  is  a  form 
of  energy,  and  that  it  is  a  kind  of  motion.     It  is  not  our  pur- 
pose to  enter  into  the  different  theories  regarding  heat,  but 
as  much  of  the  generally  accepted  theory  will  be  given  as 
will  be  necessary  to  make  clear  the  principles  that  are  to 
follow. 

78.  In  Art.  3  it  was  stated  that  bodies  are  composed  of 
molecules.     Notwithstanding  the  extreme  minuteness  of  the 
molecules,  they  play  a  very  important  part  in  the  modern 
theory  of  heat.      Each  molecule  attracts  the  molecules  sur- 
rounding it  in  a  manner  similar  to  the  attraction  between 
earth  and  bodies  near  its  surface,  only  with  an  immensely 
greater  force  in  proportion  to  their  sizes.     Without  going 
into  any  theory  regarding  the  precise  nature  of  heat,  it  will 
be  taken  for  granted  that  each  and   every  molecule  has  a 
rapid  vibratory  motion  to  and  fro,  and  that  the  molecules  are 
kept  from  getting  beyond  a  certain  distance  from  one  another 
by  the  attractive  force  between  them.     This  attractive  force 
is  called   cohesion;  without  it,   everything  throughout  the 
universe  would  instantly  crumble  into  the  finest  dust. 

In  Art.  15  it  was  stated  that  the  molecules  are  supposed 
to  be  round ;  it  is  likewise  supposed  that  they  are  at  a  con- 
siderable distance  apart,  compared  with  their  diameters. 
When  heat  is  applied  to  a  body,  the  number  of  the  vibrations 


42  PHYSICS.  §  4 

is  increased  proportionally  to  the  amount  of  heat  supplied. 
In  consequence  of  this  increase,  the  distance  through  which  a 
molecule  moves  is  increased,  and  the  force  of  cohesion  that 
binds  the  molecules  together  is  lessened.  If  enough  heat  is 
added  to  a  solid,  the  force  of  cohesion  is  so  far  overcome  that 
the  body  melts.  If  heat  is  supplied  in  sufficient  quantity,  the 
melted  body  becomes  a  vapor,  and  so  long  as  it  is  kept  at 
this  temperature  the  force  of  cohesion  has  no  effect,  in  con- 
sequence of  the  number  of  vibrations  having  been  so  far 
increased  and  the  distance  between  any  two  molecules  hav- 
ing become  too  great  for  the  force  of  cohesion  to  act.  If  the 
vapor  is  cooled,  the  number  of  vibrations  and  also  the  dis- 
tance between  any  two  molecules  will  decrease ;  the  force  of 
cohesion  begins  to  act  and  the  body  becomes  a  liquid.  If 
cooled  further  and  a  sufficient  quantity  of  heat  is  removed 
(in  other  words,  if  the  number  of  vibrations  is  so  far  decreased 
that  the  molecules  are  comparatively  near  together),  the  body 
becomes  a  solid  and  remains  so  until  the  temperature  is  again 
increased  to  the  melting  point. 

79.  If  a  body  is  heated  and  brought  near  the  hand,  the 
sensation  of  warmth  is  felt;  if  heat  is  removed  from  this 
same  body  and  it  is  again  brought  near  the  hand,  the  sensa- 
tion of  cold  is  felt.  The  heat  that  thus  manifests  itself  is 
called  sensible  heat,  because  any  change  from  any  state  to  a 
hotter  or  colder  state  is  indicated  at  once  by  the  sense  of 
feeling.  The  more  sensible  heat  a  body  possesses,  the  hotter 
it  is ;  the  more  sensible  heat  is  taken  away  from  it,  the  colder 
it  is. 


TEMPERATURE. 

8O.  The  most  common  heat  measure  we  have  is  that 
which  we  gain  by  means  of  the  sensation  of  warmth  it 
produces.  According  to  the  character  of  this  sensation  a 
body  is  said  to  be  cold,  warm,  or  hot.  These  terms  all 
refer  to  the  power  that  one  body  has  of  communicating 
heat  to  other  bodies.  The  measure  of  this  power  is  termed 


§  4  PHYSICS.  43 

temperature,  which  may  be  more  exactly  embodied  in  the 
following"  definition:  The  temperature  of  a  body  is  a  measure 
of  tJie  intensity  of  its  licat,  and  is  further  defined  as  the 
thermal  state  of  a  body  considered  with  reference  to  its  power 
of  communicating  heat  to  other  bodies. 

8 1 .  For  scientific  purposes,  the  sensations  are  not  suffi- 
ciently accurate  methods  of  measuring  temperature ;  accord- 
ingly,  temperature  is  usually  measured  by  certain  of    the 
effects  produced  by  heat.     Among  these,  one  of  the  most 
convenient  is,  that  most  bodies  expand  with  a  rise  of  tem- 
perature.    This  expansion  is  distinctly  perceptible  in  solids; 
it  occurs  to  a  greater  extent  in  liquids,  and  most  of  all  with 
gases.     For  the  general  purposes  of  temperature  measure- 
ment,  mercury  and  alcohol  are  the  most  convenient  sub- 
stances,   the  former  because   it  boils  only  at  a  very  high 
temperature,  and  the  latter  because  it  does  not  solidify  at 
the  greatest  known  cold  produced  by  ordinary  means.     One 
of  these  liquids,  enclosed  in  a  suitable  vessel,  constitutes  the 
temperature-measuring  instrument  termed  a  thermometer. 

82.  In  constructing1  a  thermometer,  a  bulb  is  blown  at 
one  end  of  a  glass  tube  of  a  very  narrow  bore ;  the  bulb  and 
a  portion  of  this  tube  are  next  filled  with  carefully  purified 
mercury;    this  is  boiled,   and  thus  all  air  and  moisture  is 
driven  out  of  the  tube;  the  open  end  is  then  hermetically 
sealed  by  fusing"  the  glass  itself.     The  bulb  and  a  portion  of 
the  tube  are  thus  filled  with  mercury  and  the  remainder  of 
the  tube  is  a  vacuum,  save  for  the  presence  of  a  minute 
quantity  of  mercury  vapor.     On  heating  the  bulb  of  this 
instrument  the  mercury  expands  and  rises  considerably  in 
the   stem.       Heat   has   a   tendency   to   so    distribute   itself 
throughout  any  body  or  series  of  bodies  as  to  make  them 
of  the  same  temperature;  consequently,  if  the  thermometer 
is  placed  in  contact  with  the  body  whose  temperature  it  is 
desired  to  measure,  a  redistribution  of  heat  occurs  until  the 
two  are  at  the  same  temperature.     That  is  to  say,  if  the  body 
is  the  colder  it  receives  heat  from  the  thermometer,  and  if 


44 


PHYSICS. 


it  is  hotter  it  yields  heat  to  the  thermometer,  until  the  tem- 
perature of  the  two  is  the  same.  The  two  being  in  efficient 
contact,  this  stage  is  indicated  by  the  mercury  becoming 
stationary  in  the  thermometer  tube.  Now  the  mercury  is 
constant  for  any  one  temperature;  therefore,  to  register 
^==^  temperature,  it  is  only  necessary  to  have,  further, 
a  scale  or  series  of  graduations  attached  to  the 
stem  of  the  instrument,  by  which  the  temperature 
may  always  be  read. 

83.  In  Fig.  14  is  shown  a  mercurial  thermom- 
eter with  two  sets  of  graduations  on  it.     The  one 
on  the  left,   marked  F,  is  the  Fahrenheit  ther- 
mometer, so  named  after  its  inventor,  and  is  the 
one  commonly  in  use  in  this  country  and  in  Eng- 
land;   the    one    on    the    right,  marked    C,    is  the 
centigrade  thermometer,  and  is  used  by  scientists 
throughout  the  world  on  account  of  the  graduations 
being  better  adapted  for  calculations. 

84.  In   graduating   thermometers   two  fixed 
points  of  temperature  are  almost  universally  em- 
ployed.    These  are  the  temperatures  of  melting 
ice,  and  of  the  steam  from  boiling  water.    Certain 
simple  precautions  being  taken,  these  temperatures 
are  always  constant.      It  is,  in  addition,  necessary 
to  graduate   the  thermometers,  so  as  to  register 
temperatures    intermediate    between    these    two 
points,   and   also   below   and  above  them.      The 
most  convenient  system  of  graduation  is  that  of 
Celsius,  known  as  the  centigrade  scale.     In  this 

scale  the  distance  between  the  melting  point  of  ice  (or  the 
freezing  point  of  water,  as  it  is  commonly  called)  and 
the  temperature  of  steam  at  atmospheric  pressure  is  divided 
into  100  equal  graduations  or  degrees.  (This  latter  tem- 
perature is  more  commonly  described  as  being  that  of  the 
boiling  point  of  water. )  The  freezing  point  is  called  zero, 
or  0°,  and  the  boiling  point  100°.  Degrees  of  the  same 


FIG.  14. 


§4  PHYSICS.  45 

value  are  carried  above  and  below  the  boiling  and  freezing- 
points,  respectively;  the  temperature  below  0°  is  considered 
negative,  and  is  counted  downwards  from  zero;  thus,  10° 
below  freezing  point  is  — 10°,  and  so  on.  Degrees  above  the 
boiling  point  are  simply  counted  upwards  from  zero;  thus, 
10°  above  the  boiling  point  is  110°,  and  so  on. 

85.  In  graduating  Fahrenheit  thermometers,  the  distance 
between  the  freezing  and  boiling  points  is  divided  into  180 
equal  parts,  and  degrees  of  the  same  size  are  carried  above 
and   below    the   boiling    and   freezing   points.      Fahrenheit 
assumed  that  the  greatest  cold  attainable  was  32°  below  the 
freezing  point,  and  accordingly  took  that  point  as  his  zero 
and  reckoned  from  it  upwards.      The  freezing  point  thus 
became  32°  F.  and  the  boiling  point  32  -f  180  =  212°.   Degrees 
below  the   Fahrenheit  zero  are  reckoned  downwards  from 
zero  and  are  considered  as  negative. 

In  Germany  and  Russia  temperature  is  reckoned  on  the 
Reaumur  scale,  in  which  the  freezing  point  is  0°  and  the 
boiling  point  is  80°. 

86.  Of  these  three  thermometers,  the  centigrade  is  used 
the  most,  but  since  the  Fahrenheit  instrument  is  the  one  in 
general  use  in  this  country,  all  temperatures  in  this  section 
will  be  understood  to  be  in  Fahrenheit  degrees,  unless  other- 
wise stated. 

87.  It  is  frequently  necessary  to  be  able   to  compare 
these  scales,  and  to  translate  temperatures  from  any  one  into 
another.      For  example,  what  would  80°  C.  be  on  the  Fah- 
renheit scale  ? 

Since  the  number  of  degrees  between  the  freezing  point 
and  the  boiling  point  on  the  centigrade  scale  is  100,  and  on 
the  Fahrenheit  180,  it  is  evident  that,  if  F  =  number  of 
degrees  Fahrenheit,  and  C  —  number  of  degrees  centigrade, 

F  :  C  ::  180  :  100,  or  F  =  }£ £  C  =  f  C. 

C   100/T   5     P 

u   —       --  r   —  --  r. 


46  PHYSICS.  §  4 

Therefore,  to  change  centigrade  temperatures  into  their 
corresponding  Fahrenheit  values: 

Rule. — Multiply  the  temperature,  centigrade,  by  \  and  add 
32° ;  the  result  ivill  be  the  temperature,  Fahrenheit. 

To  change  Fahrenheit  temperatures  into  their  correspond- 
ing centigrade  values: 

Rule. — Subtract  32°  from  the  temperature,  Fahrenheit, 
multiply  by  f,  and  the  result  ivill  be  the  temperature,  centi- 
grade. 

Expressing  these  two  rules  by  means  of  formulas, 

Let  /,.  =  temperature  centigrade; 

tf  =  temperature  Fahrenheit. 
Then,  tf  =  £/,  +  32°,  (19.) 

and  4.  =  -§(/>- 32°).  (2O.) 

EXAMPLE  1.— Change  (a)  100°'  C.,  (b)  4°  C.,  and  (c)  -40°  C.  into  Fah- 
renheit temperatures. 

SOLUTION.—  (a)  tf  =  f  4  +  32  =  f  XlOO  +  32  =  212°  F.     Ans. 

(b)  tf  =  |x4  +  32  =  39.2°  F.     Ans. 

(c)  tf  =  f  X  -40  +  32  =  -40°  F.     Ans. 

EXAMPLE  2.— Change  (a)  60°  F.,  (b)  32°  F.,  and  (c)  —20°  F.  into  their 
corresponding  centigrade  temperatures. 

SOLUTION.—  (a)  tc  =  (/r-32)|  =  (60-32)f  =  15|°  C.     Ans. 

(b)  4  =  (32-32)f  =  0°  C.     Ans. 

(c)  4  =  (-  20  -  32)f  =  -  28f  °  C.     Ans. 

88.  The  absolute  temperature  is  the  temperature  meas- 
ured above  the  point  of  absolute  zero  (see  Art.  64).  Hence, 
on  the  Fahrenheit  scale  the  absolute  temperature  T  is 
460°  -{- 1°  when  t  —  the  ordinary  temperature,  and  is  above 
zero.  If  t°  is  below  zero,  its  value  is  negative,  and  the 
absolute  temperature  Tis  460° -j-(— 1°)  =  460°  —  t°. 

Throughout  this  section,  where  temperatures  are  men- 
tioned, t  will  denote  the  ordinary  temperature  indicated  by 
the  thermometer,  and  T,  the  absolute  temperature. 

EXAMPLE. — What  are  the  absolute  temperatures  corresponding  to 
the  Fahrenheit  temperatures  212°,  32°,  and  —39.2°? 


§  4  PHYSICS.  47 

SOLUTION.—  460°  +  212°  =  T  =  672°  F.  Ans. 
460°  +  32°  =  T  =  492°  F.  Ans. 
460°  -39.2°  =  T  =  420.8°  F.  Ans. 

The  absolute  temperature  on  the  centigrade  scale  is 
T  =  273i°  +  /°  when  t°  is  above  zero,  or  T  =  273^°  - 1° 
when  t°  is  below  zero. 

EXAMPLE. — What  are  the  absolute  temperatures  corresponding  to 
100°  C.,  4°  C.,  and  —40°  C.? 

SOLUTION.—  2731°  +100°  =  T  =-.  373i°  C.  Ans. 
2731°  +  4°  =  T  =  2771°  C.  Ans. 
2731°  -  40°  =  T  =  2331°  C.  Ans. 


EXAMPLES  FOR  PRACTICE. 

89.      1.     What  are  the  absolute  temperatures  corresponding  to 
(a)  120°  R.?  (b)  120°  C.?  (c)  120°  F.?  f  (a}    338j-  R> 

Ans.  I   (6)     393|°C. 
[  (c)      580°  F. 

2.     Change  —  10°  R.  to  the  corresponding  Fahrenheit  and  centigrade 
readings.  Ans.  91°  F. ;  -  12|°  C. 


OF    BODIES    BY    HEAT. 

90.  The  volume  of  any  body — solid,  liquid,  or  gaseous — 
is  always  changed  if  the  temperature  is  changed;  nearly  all 
bodies  expand  whin  heated,  and  contract  when  cooled.     In 
solids  having  definite  figures,  the  expansion  may  be  consid- 
ered in  three  ways,  according  to  the  conditions. 

1.  The  expansion  in  one  direction,  as  the  elongation  of  an 
iron  bar;  this  is  called  linear  expansion. 

2.  Surface  expansion,  where  the  area  is  increased. 

3.  Cubical  expansion,  where  the  increase  in  the  whole 
volume  is  considered. 

91.  In  Fig.  15  is  shown  an  apparatus  for  exhibiting  the 
linear  expansion  of  a  solid  body.     A  metal  rod  A  is  fixed  at 
one  end  by  a  screw  B,  the  other  end  passing  freely  through 


48 


PHYSICS. 


the  eye  6",  held  in  the  post,  and  pressing  against  the  short 
arm  of  the  indicator  F.  The  rod  is  heated  as  shown,  and  its 
elongation  causes  the  indicator  to  move  along  the  arc  D  E. 

An  illustration  of  surface  expansion  is  afforded   nearly 
every  day  in    machine   shops,    particularly   in   locomotive 


FIG.  15. 

shops,  where  piston  rods,  crankpins,  etc.  are  shrunk  in,  and 
tires  are  shrunk  on  their  centers.  In  shrinking  a  tire  on,  it 
is  bored  a  little  smaller  than  the  wheel  center.  The  tire  is 
then  heated  until  the  area  of  its  circumference  is  expanded 
enough  to  allow  it  to  slide  over  the  wheel  center.  On  being 


FIG.  16. 


cooled  with  water,  the  tire  contracts,  and  tends  to  regain  its 
original  area,  but  is  prevented  by  reason  of  the  wheel  center 
being  a  trifle  larger.  This  causes  the  tire  to  hug  the  center 
with  immense  force,  and  prevents  it  from  coming  off. 


§  4  PHYSICS.  49 

Cubical  expansion  may  be  illustrated  by  means  of  a  Grave- 
sandes'  ring.  This  consists  of  a  brass  ball  «,  Fig.  16,  which 
at  ordinary  temperatures  passes  freely  through  the  ring  m, 
of  very  nearly  the  same  diameter.  When  the  ball  is  heated 
it  expands  so  much  that  it  will  no  longer  pass  through  the 
ring. 

The  expansion  of  liquids  is  clearly  shown  in  the  mercurial 
and  alcohol  thermometers.  The  expansion  of  gases  was 
treated  to  some  extent  in  Arts.  56  to  72. 


92.  Coefficient  of  Expansion. — Suppose  that  the  tem- 
perature of  the  metal  rod  shown  in  Fig.  15  was  32°  F.  before 
heating,  and  that  its  length  at  that  temperature  is  exactly 
10  feet.  It  is  found  that  when  the  temperature  is  raised  1°, 
or  to  33°,  the  bar  is  10  f eet  -|-  TYOT  mcn  l°ng-  The  linear 
expansion  is  then  (10  feet  +  TTTO-  inch)  — 10  feet  =  y^V o"  inch, 
and  the  ratio  between  this  expansion  and  the  original  length 
of  the  bar  is 

:  10X12  =  ..        *    .,_  =  .000006944. 


1,200  1,200X120 

For  every  increase  of  temperature  of  1°,  this  rod  elongates 
.000006944  of  its  length.  This  number  .000006944,  which  is 
equal  to  the  expansion  of  the  rod  for  1°  rise  of  temperature 
divided  by  the  original  length,  is  called  the  coefficient  of 
linear  expansion.  Had  the  temperature  of  the  rod  been 
increased  100°  instead  of  1°,  the  amount  of  elongation  would 
have  been  .000006944x100  =  .0006944  of  its  length,  or 
.0006944X120  =  .083328  inch,  or  TV  inch. 

93.  The  table  following  contains  the  coefficients  of 
expansion  for  a  number  of  solids,  mercury,  and  alcohol,  and 
the  average  cubical  expansion  of  gases.  No  liquids  are  given 
except  mercury  and  alcohol,  for  the  reason  that  the  coeffi- 
cient of  expansion  for  liquids  is  different  at  different  tem- 
peratures. 


50 


PHYSICS. 


TABLE  2. 
TABLE  OF  COEFFICIENTS  OF  EXPANSION. 


Name  of  Substance. 

Linear 
Expansion. 

Surface 
Expansion. 

Cubical 
Expansion. 

Cast  iron  

.00000617 

.00001234 

00001  8.50 

Copper 

00000955 

00001910 

00002864 

Brass  

00001037 

00002074 

00003112 

Silver  

00000690 

00001390 

00002070 

Bar  iron  

.00000686 

00001372 

00002058 

Steel  (tin  tempered)  .  .  . 
Steel  (tempered)  .... 

.00000599 
.  00000702 

.00001198 
.00001404 

.00001798 
.00002106 

Zinc  

00001634 

.  00003268 

.00004903 

Tin 

00001410 

00002820 

00003229 

Mercury  
Alcohol  

.00003334 
.00019259 

.00006668 
.00038518 

.00010010 

.00057778 

Gases  

.00203252 

94.     Let     L    =  length  of  any  body ; 

/  =  amount  of  expansion  or  contraction  due 
to  heating  or  cooling  the  body ; 

A    —  area  of  any  section  of  the  body; 

a  —  increase  or  decrease  of  area  of  the  same 
section  after  heating  or  cooling  the  body ; 

V  —  volume  of  the  body ; 

v  —  increase  or  decrease  in  volume  due  to 
heating  or  cooling  the  body; 

C\  —  coefficient  of  expansion  taken  from  col- 
umn 1,  Table  2; 

C3  —  coefficient  taken  from  column  2,  Table  2 ; 

6"3  =  coefficient  taken  from  column  3,  Table  2 ; 

/  —  difference  in  degrees  of  temperature 
between  the  original  temperature  and 
the  temperature  of  the  body  after  it  has 
been  heated  or  cooled. 


§  4  PHYSICS.  51 

Then,  /  =  LC,t. 

a  =  A  C  t. 


v  =   VC\t. 

EXAMPLE.  —  How  much  will  a  bar  of  untempered  steel  14  feet  long 
expand,  if  its  temperature  is  raised  80°  ? 

SOLUTION.  —  Since  only  one  dimension  is  given,  that  of  length,  linear 
expansion  only  can  be  considered.  From  Table  2,  the  coefficient  of 
linear  expansion  per  unit  of  length  for  a  rise  in  temperature,  of  1°  is 
found  to  be  .00000599  for  untempered  steel.  Hence,  using  formula  21, 
/=  L  dt,  and  substituting,  14  X  .  00000599  X  80  =  .0067088  foot,  or 
.0067088  X  12  =  .0805056  inch.  Ans. 

95.  This  expansion  seems  very  small,  but  in  engineer- 
ing1 constructions,  when  long  pieces  are  rigidly  connected,  it 
must  »be  taken  into  account.  If  the  cross-section  of  the 
above  bar  was  "Z  inches  square,  and  the  bar  was  fitted  tightly 
between  two  supports,  an  expansion  of  the  above  amount 
would  exert  a  pressure  against  the  supports  of  about  58,000 
pounds. 

Suppose  that  an  iron  rod  1^  inches  in  diameter  and  100  feet 
long  were  used  as  a  tie-rod  in  constructing  a  bridge  ;  that  it 
was  put  in  place  and  securely  fastened  to  two  rigid  supports 
diiring  a  warm  day  in  summer  when  the  temperature  in  the 
sunlight  was,  say,  110°.  On  a  cold  day  in  winter  when  the 
thermometer  registered  zero,  the  amount  that  the  bar  would 
tend  to  shorten  owing  to  this  change  in  temperature  would 
be,  substituting  these  values  in  formula  21, 

.00000686X100X110  =  .07546  foot  =  .90552  inch. 

If  this  rod  were  rigidly  secured  so  that  it  could  neither 
stretch  nor  shorten,  it  would  then  exert  a  pull  on  the  sup- 
ports of  about  33,  400  pounds. 

EXAMPLE  1.  —  The  wheel  center  of  a  locomotive  driver  is  turned  to 
exactly  50  inches  in  diameter.  If  the  steel  tire  be  bored  49.94  inches  in 
diameter,  to  what  temperature  must  the  tire  be  raised  in  order  that  it 
may  be  easily  shoved  over  the  center  ?  Assume  that  the  diameter  of 
the  tire  is  expanded  to  y-gVu  °f  an  inch  larger  than  the  center,  and  that 
the  original  temperature  is  60°. 


52  PHYSICS.  §  4 

SOLUTION.  —For  this  case,  formula  22  may  be  used.  The  original 
diameter  of  the  tire  is  49.94  inches,  and  it  is  to  be  increased  to  50.001 
inches.  The  area  of  a  circle  49.94  inches  in  diameter  is  1,958.79  square 
inches.  Area  of  circle  50.001  inches  in  diameter  is  1,963.58  square 
inches.  The  difference  between  them  is  1,963.58-1,958.79  =  4.79 
square  inches  =  a  in  formula  22.  Hence,  since  C2  =  .00001198,  and 
A  =  1,958.79,  substitute  these  values  in  a  =  A  C2  /,  and  4.79  =  1,958.79 

4.  7Q 

X.  00001198  x/  =  .023466  /.     Therefore,   /  =    ^^  =  204.125°,   and 
204.  125°  +  60°  =  264.125°.     Ans. 

NOTE.  —  Owing  to  the  form  of  the  equation  here  denoted  by  formula 
22,  and  to  the  manner  in  which  the  coefficients  C2  were  determined, 
this  example  may  be  more  easily  solved  by  means  of  formula  21. 
Thus,  regard  the  diameter  as  a  linear  dimension  and  apply  formula  SI. 
Increase  in  diameter  =  /  =  50.001  —  49.94  =  .061  inches.  L  =  49.94 
and  Ci  =  .00000599.  Substituting,  .061  =  49.94  X  .00000599  X  /,  or, 


49.94 

The  slight  difference  in  the  two  answers  is  immaterial,  and  is  to  be 
expected. 

EXAMPLE  2.  —  What  is  the  decrease  in  volume  of  a  copper  cylinder 
30  inches  long  and  22  inches  in  diameter,  if  cooled  from  212°  to  0°,  the 
measurement  being  taken  at  a  temperature  of  70°  ? 

SOLUTION.  —  212°  —  70°  =  142°  =  the  increase  in  temperature  above 
70°.  Use  formula  23,  v  =  V  C3  1. 

V  =  222  X  .7854  X  30  =  11,404  cu.  in. 
v  =  11,404  X  .00002864  X  142  =  46.38  cu.  in. 
11,404  +  46.38  =  11,450.38  cu.  in.  =  the  volume  at  212°. 
70°  —  0°  =  70°  =  the  difference  in  temperature. 

V  =  11,404  X  .00002864  X  70  =  22.86  cu.  in.,  nearly. 
46.38  +  22.86  =  69.24  cu.  in.     Ans. 

96.  The  bars  of  a  furnace  must  not  be  fitted  tightly  at 
their  extremities,  but  must  be  free  at  one  end,  otherwise  in 
expanding  they  will  split  the  masonry. 

In  laying  the  rails  on  railroads,  a  small  space  is  left  between 
the  successive  rails;  for,  if  they  touched,  the  force  of  expan- 
sion would  cause  them  to  curve.  Water  pipes  are  fitted  to 
one  another  by  means  of  telescope  joints,  which  allow  room 
for  expansion;  so,  also,  are  steam  pipes,  by  means  of  the 
so  called  expansion  joints.  If  a  glass  vessel  is  heated  or 
cooled  too  rapidly,  it  cracks,  especially  if  it  is  thick;  the 
reason  for  this  is  that  since  glass  is  a  poor  conductor  of  heat, 


§  4  PHYSICS.  53 

the    sides    become    unequally    heated,    and,    consequently, 
unequally  expanded,  which  causes  a  fracture. 

97.  It  will  be  found  on  trial  that  the  three  preceding 
formulas  will  not  work  back  ;  i.  e.  ,  if  the  length  of  a  bar  after 
it  has  been  heated  is  found  by  formula  21,  and  an  attempt 
is  made  to  reduce  the  bar  to  its  original  length  by  again 
applying  formula  21  and  substituting  for  t  the  same  value 
as  in  the  first  case,  the  value  obtained  for  /  will  be  slightly 
different  in  the  two  cases.  The  difference,  however,  is  so 
slight  that  it  is  neglected  in  practice.  If,  however,  the 
student  desires  to  obtain  exactly  the  same  result  in  both 
cases,  he  must  use  the  following  more  cumbersome  formula, 
in  which  t^  /.2,  /15  /2  are,  respectively,  the  original  and  final 
temperatiires,  the  original  and  final  lengths,  and  Cl  has  the 
same  value  as  in  formula  21  : 


This  formula  is  always  used  when  calculating  the  expan- 
sion of  gases  by  substituting  v^  z/a,  and  -^^  for  /,,  /2,  and  Clt 
respectively.  The  result  obtained  will  be  exactly  the  same 
as  those  obtained  by  formula  12.  For,  substituting  the 
values  as  directed,  the  formula  becomes 

492      /  -  32 


NX  7; 

492  +  /,-32       * 
492 


98.  Although,  as  stated  previously,  the  expansion  of 
solids  and  liquids  is  very  nearly  uniform  throughout  all 
ranges  of  temperature,  water  is  a  marked  exception  to  the 
general  rule.  If  water  is  cooled  down  from  its  boiling  point, 
it  continually  contracts  until  it  reaches  39.2°  F.,  when  it 
begins  to  expand,  until  it  freezes  at  32°  F.  On  the  other 
hand,  if  water  at  32°  F.  is  heated,  it  contracts  until  it  reaches 
39.2°  F.,  when  it  commences  to  expand.  Therefore,  the 


54 


PHYSICS. 


density  of  water  is  greatest  where  this  change  occurs.  The 
importance  of  this  exception  is  seen  in  the  fact  that  ice  forms 
on  the  surface  of  the  water,  since  it  is  lighter  than  the  warmer 
body  of  water  lying  at  varying  depths  below  it.  Were  it  not 
for  this  fact,  all  the  large  bodies  of  water  would  freeze  solid 
and  would  so  affect  the  earth's  climate  that  it  would  be  unin- 
habitable. t  The  coefficient  of  expansion  of  water  is  such  a 
very  changeable  quantity  (varying  with  the  temperature) 
that  a  special  table  is  necessary. 

99.  The  effect  of  heat  on  the  expansion  of  gases  was 
treated  in  Arts.  61  to  67,  and  will  not  be  repeated  here. 
It  should  be  stated,  however,  that  the  constant  .37052  used 
in  formulas  14  and  15  has  that  value  for  air  only.  For 
other  gases  it  varies.  If  the  value  of  this  constant  for  any 
gas  be  represented  by  R,  formula  15  becomes 

pV  =  RWT.  (25.) 

1  GO.  The  value  R  for  several  gases  is  given  in  the  fol- 
lowing table: 

TABLE  3. 


Gas. 

Volume  of  1  Lb. 
at  32°  F.   and  a 
Tension  of  1  At- 
mosphere    (14.7 
Lb.  per  Sq.  In.). 
Cubic  Feet. 

Weight  of  1  Cu. 
Ft.  at  32°  F.  and  a 
Tension  of  1  At- 
mosphere    (14.7 
Lb.  per  Sq.  In.). 
Pounds. 

R. 

Air 

12  3880 

08073 

.  37052 

Oxvtren 

11  2056 

.08925 

.33552 

Nitrogen 

12  7226 

07860 

.38143 

Hydrogen  

178.8910 

.00559 

5.34946 

EXAMPLE  1. — What  is.  the  volume  of  3  ounces  of  hydrogen  gas  having 
a  tension  of  20  pounds  per  square  inch  and  a  temperature  of  80°  ? 

SOLUTION. —    3  ounces  =  T8¥  of  a  pound.     Since  /  =  80° ,    T  —  460 
+  80  =  540°,  R  =  5.34946  from  Table  3.     Hence,  by  formula  25, 
p  V  =  R  W  T,  or  20  V  =  5.34946  X  A  X  540  =  541.6328, 

and  V  —  - — o7r — •  =  27.08164  cu.  ft. ;  say,  27.082  c.u.  ft.     Ans. 


§  4  PHYSICS.  55 

EXAMPLE  2. — What  is  the  weight  of  10  cubic  feet  of  oxygen  having 
a  tension  of  1  atmosphere  and  a  temperature  of  60°  ? 
SOLUTION. — By  formula  25, 

p  V  =  A'  W  T,  or  10  X  14.7  =  .33553  X  H'X  520. 

147 

Hence,  147  =  174.4704  W,  and  W  =       ". '       .-  .84255  Ib.     Ans. 

1  74. 4  4  U4 

In  Table  2  the  coefficient  of  expansion  for  gases  was 
given  as  .  00203252 ;  this  is  the  fraction  ^i^  reduced  to  a  deci- 
mal. This  value  of  the  coefficient  of  expansion  is  very 
nearly  the  same  for  all  gases,  particularly  for  those  that  are 
very  difficult  to  liquefy. 


EXAMPLES  FOll  PRACTICE. 

1O1.      Solve  the  following: 

1.  (a)  How  much  will  an  iron  tie-rod  60  feet  long  expand  when  the 
temperature  is  raised  from  40°  to.  110°  ?     (b)  Calculate  also  by  formula 
24.     (c)  What  is  the  difference  in  the  two  results  ?       (  ta^      345744  in 

Ans.  •{  (b)     .845725  in. 
[  (0      .000019  in. 

2.  To  what  temperature  must  a  steel  tire  of  59.93  inches  internal 
diameter  be  raised  in  order  that  its  diameter  may  be  60.0015  inches  ? 
Original  temperature  =  71°.  Ans.  270°. 

3.  What  is  the  volume  of  .68  pound  of  nitrogen  gas  having  a  ten- 
sion of  20  pounds  per  square  inch  and  a  temperature  of  345°  ? 

Ans.  10.44  cu.  ft. 


HEAT    PROPAGATION. 

1O2.  Conduction  of  Heat. — Conduction  is  the  slow 
progress  of  the  vibratory  motion  from  places  of  higher  to 
places  of  lower  temperature  in  the  same  body.  Different 
bodies  possess  very  different  conducting  powers,  the  good 
conductors  being  those  in  which  conduction  is  most  rapid, 
and  the  bad  conductors  those  in  which  it  is  very  slow. 
Experiments  have  established  a  numerical  comparison  of  the 
conducting  powers  of  many  bodies.  Representing  the  con- 
ducting power  of  silver  by  100,  the  following  table  shows 


56 


PHYSICS. 


the  conducting1  power  of  a  number  of  other  metals  as  deter- 
mined by  Wiedemann  and  Franz : 


TABLE  4. 


Silver 100.0 

Copper 73 . 6 

Gold 53.2 

Brass 23.1 

Zinc 19.0 

Tin.,  14.5 


Iron 11.9 

Steel 11.6 

Lead 8.5 

Platinum 8.4 

Rose's  alloy 2.8 

Bismuth.  .  1.8 


1O3.  As  a  class,  the  metals  are  the  best  conductors;  the 
fluids,  both  liquid  and  gaseous,  are  very  poor  conductors, 
their  conducting"  power  being  hardly  appreciable.  Water, 
for  example,  can  •  be  made  to  boil  at  the  top  of  a  vessel, 
while  a  cake  of  ice  is  fastened  within  a  few  inches  of  the 
surface.  If  thermometers  are  placed  at  different  depths, 
while  water  boils  at  the  top,  it  is  found  that  the  conduction 
of  heat  downwards  is  very  slight. 

Organic  substances  conduct  heat  poorly.  This  enables 
trees  to  withstand  great  and  sudden  changes  in  the  atmos- 
phere without  injury.  The  bark  is  a  poorer  conductor  than 
the  wood  beneath  it.  Cotton,  wool,  straw,  bran,  etc.  are  all 
poor  conductors.  Rocks  and  earth  are  poorer  conductors, 
the  less  dense  and  homogeneous  the  mass ;  hence,  the  length 
of  time  required  for  the  sun's  rays  to  penetrate  the  earth. 
The  mean  highest  temperature  of  the  air  near  the  ground  in 
Central  Europe  is  in  the  month  of  July,  but  at  a  depth  of 
from  25  to  28  feet  in  the  earth  the  highest  temperature  occurs 
in  the  month  -of  December. 


1O4.  Dynamical  Theory  of  Heat. — Before  going  any 
further,  it  will  be  convenient  to  explain  here  the  view  now 
generally  taken  as  to  the  mode  in  which  heat  is  propagated. 
On  this  subject,  it  is  stated  in  Ganot's  Physics,  * '  A  hot  body 
is  one  whose  molecules  are  in  a  state  of  vibration.  The 
higher  the  temperature  of  a  body,  the  more  rapid  are  these 


§  4  PHYSICS.  57 

vibrations,  and  a  diminution  in  temperature  is  but  a  dimin- 
ished rapidity  of  the  vibration  of  the  molecules.  The 
propagation  of  heat  through  a  bar  is  due  to  a  gradual  com- 
munication of  this  vibratory  motion  from  the  heated  part  to 
the  rest  of  the  bar.  A  good  conductor  is  one  that  readily 
takes  up  and  transmits  the  vibratory  motion  from  molecule 
to  molecule,  while  a  bad  conductor  is  one  that  takes  up  and 
transmits  the  motion  with  difficulty.  But  even  through  the 
best  of  the  conductors,  the  propagation  of  this  motion  is 
comparatively  slow.  How,  then,  can  be  explained  the 
instantaneous  perception  of  heat  when  a  screen  is  removed 
from  a  fire,  or  when  a  cloud  drifts  from  the  face  of  the  sun  ? 
In  this  case,  the  heat  passes  from  one  body  to  another  with- 
out affecting  the  temperature  of  the  medium  that  transmits 
it.  In  order  to  explain  these  phenomena,  it  is  imagined  that 
all  space,  the  space  between  the  planets  and  the  stars,  as 
well  as  the  interstices  in  the  hardest  crystal  and  the  heaviest 
metal — in  short,  matter  of  any  kind — is  permeated  by  a 
medium  having  the  properties  of  matter  of  infinite  tenuity, 
called  ether.  The  molecules  of  a  heated  body  being  in  a 
state  of  intensely  rapid  vibration,  communicate  their  motion 
to  the  ether  around  them,  throwing  it  into  a  system  of  waves, 
which  travel  through  space  and  pass  from  one  body  to 
another  with  the  velocity  of  light.  When  the  undulations  of 
the  ether  reach  a  given  body,  the  motion  is  given  up  to  the 
molecules  of  that  body,  which,  in  their  turn,  begin  to 
vibrate ;  that  is,  the  body  becomes  heated.  The  process  of 
this  motion  through  the  ether  is  termed  radiation,  and  what 
is  called  a  ray  of  heat  is  merely  one  series  of  waves  moving 
in  a  given  direction. 

HEAT  MEASUREMENT. 


THERMAL  UNITS. 

1O5.  Quantity  of  Heat. — Temperature  is  not  a  measure 
of  quantity  of  heat,  for  a  thermometer  would  indicate  the 
same  temperature  both  in  a  vessel  containing  a  pint,  and 
one  containing  a  gallon,  of  boiling  water,  although  it  is 


58  PHYSICS.  §  4 

evident  that  the  one  must  contain  eight  times  as  much  heat  as 
the  other;  further,  to  raise  the  gallon  of  water  to  the  boiling 
point,  eight  times  the  amount  of  heat  necessary  to  similarly 
raise  the  pint  is  required.  This  leads  us  to  the  mode  of 
measuring  and  indicating  quantity  of  heat.  Quantity  of 
heat  is  measured  by  the  amount  necessary  to  raise  a  certain 
weight  of  some  body  from  one  temper  atiire  to  another  fixed 
temperature. 

1O6.     There  are   three  units  in  use  for  measuring  the 
quantity  of  heat: 

1.  The   British,  thermal    unit,   which    represents   the 
amount  of  heat  required  to  raise  1  pound  of  water  1°  Fahren- 
heit.    Instead  of  writing  out   the  words  "British   thermal 
unit  "  in  full,  it  is  customary  to  abbreviate  them  to  B.  T.  U. 
Thus,  7  pounds  of  water  raised  15°  F. ,  would  equal  7x15 
=  105  B.  T.  U.      The  unit  of  heat  used  in  this  section  will 
be  the  British  thermal  unit. 

2.  The  thermal  unit,  which   represents  the  amount  of 
heat  that  is  necessary  to  raise  1  pound  of  water  1°  centigrade. 
Since  1°  C.   =  f  X  1°  F.,  it  follows  that  the  thermal  unit  is  f 
times  as  large  as  a  B.  T.  U.     Hence,  to  change  B.  T.  U.  into 
thermal  units,  multiply  the  number  of   B.  T.  U.  by  f.      To 
change  thermal  units  into  B.  T.  U.,  multiply  the  number  of 
thermal  units  by  f. 

3.  The  calorie,  which   represents  the  amount  of   heat 
necessary  to  raise  1  kilogram  of  water  1°  centigrade.     One 
kilogram  =  2.2    pounds    and    1°  C.   =  |Xl°  F. ;    hence,    a 
calorie  =  2.2  Xf  —  3.96  B.  T.  U.       The  calorie  is  used  in 
Germany,  France,  and  in  other  countries  using  the  metric 
system  of  weights  and  measures. 


SPECIFIC  HEAT. 

1O7.  When  equal  weights  of  two  different  substances 
having  the  same  temperature  are  placed  in  similar  vessels 
and  subjected  for  the  same  length  of  time  to  the  heat  of  the 
same  lamp,  or  are  placed  at  the  same  distance  in  front  of  the 
same  fire,  it  is  found  that  their  final  temperatures  will  differ 


g  4  PHYSICS.  59 

considerably ;  thus,  mercury  will  be  much  hotter  than  water. 
But,  as,  from  the  conditions  of  the  experiment,  they  have 
each  been  receiving  the  same  amount  of  heat,  it  is  clear  that 
the  quantity  of  heat  sufficient  to  raise  the  temperature  of 
mercury  through  a  certain  number  of  degrees  will  raise  the 
same  weight  of  water  through  a  less  number  of  degrees;  in 
other  words,  it  requires  more  heat  to  raise  a  certain  weight 
of  water  1°  than  it  does  to  raise  the  same  weight  of  mercury 
1°.  Conversely,  if  the  same  quantities  of  water  and  of  mer- 
cury at  200°  are  allowed  to  cool  down  to  the  temperature  of 
the  room,  the  water  will  require  a  much  longer  time  for  the 
purpose  than  the  mercury;  hence,  in  cooling  through  the 
same  number  of  degrees,  water  gives  up  more  heat  than 
mercury. 

The  number  of  heat  units  or  parts  of  a  heat  unit  required 
to  raise  the  temperature  of  a  given  quantity  of  any  substance 
1°  is  called  the  specific  heat  of  that  substance.  It  will  be 
seen  from  the  above  definition,  that  tJic  specific  Jieat  of  a 
substance  is  the  ratio  between  the  amount  of  heat  required  to 
raise  the  temperature  of  the  substance  1°  and  the  amount  of 
the  heat  required  to  raise  the  temperature  of  the  same  weight 
of  water  1°. 

1O8.  If  the  specific  heat  of  lead  were  given  as  .0314,  it 
would  mean  that  the  amount  of  heat  required  to  raise  a  cer- 
tain weight  of  lead  1°  would  raise  the  same  weight  of  water 
only  .0314  of  1°;  or  it  would  mean  that  .0314  B.  T.  U.  would 
raise  the  temperature  of  1  pound  of  lead  1°  F. 

EXAMPLE. — The  specific  heat  of  copper  is  .0951 ;  how  many  B.  T.  U. 
will  it  take  to  raise  the  temperature  of  75  pounds  180°? 

SOLUTION. — Since  it  takes  .0951  B.T.  U.  to  raise  1  pound  of  copper  1°, 
it  will  take  .0951  X  75  X  180  to  raise  75  pounds  180°.     Hence, 
.0951  X  75  X  180  =  1,283.85  B.T.U.     Ans. 

In  the  example  just  given,  if  it  had  been  required  to  raise 
75  pounds  of  water  180°  (that  is,  from  the  freezing  point  to 
the  boiling  point),  it  would  have  taken  75x180  =  13,500 

B.  T.  U.,and1>283'85  =  .0951  =  the  specific  heat  of  copper. 
13,500 


60 


PHYSICS. 


109.  The  following"  is  the  formula  for  finding  the  num- 
ber of  B.  T.  U.  required  to  raise  the  temperature  of  a  substance 
a  given  number  of  degrees,  or  for  finding  the  number  of 
B.  T.  U.  given  up  by  a  body  in  cooling  a  given  number  of 
degrees. 

Let  W  =  weight  of  body  in  pounds ; 

s    —  specific  heat  of   the  substance  composing  the 

body; 

/    =  original  temperature  of  body; 
/j    =  final  temperature  of  body; 

n    =  number  of  B.  T.  U.   required,  or  given  up,   in 
changing  the  temperature  of  the  body  from  t° 
to  ff. 
Then,  n  =   Wfa-t)*.  (26.) 

EXAMPLE. — A  piece  of  wrought  iron  weighing  31. 3  pounds  and  having 
a  temperature  of  900°  is  cooled  to  a  temperature  of  60° ;  how  many 
units  of  heat  does  it  give  up  ?  The  specific  heat  of  wrought  iron  is .  1 1 38. 

SOLUTION. — Apply  formula  26, /?=    lV(ti—t)s.     Substituting, 
n  =  31.3  X  (900-60)  X  .1138  =  2,992.03  B.  T.  U.     Ans. 

If  a  body  be  cooled  from  a  temperature  /  clown  to  a  tem- 
perature /,,  the  value  of  ;/  will  be  negative,  the  minus  sign 
indicating  that  the  body  was  cooled. 

110.  In  the  following  table  are  given  the  specific  heats 
of  a  number  of  substances  under  constant  pressure; 

TABLE    5. 

SOLIDS. 


Copper 0.0951 

Gold 0.0324 

Wrought  iron 0.1138 

Steel  (soft) 0.1165 

Steel  (hard) 0.1175 

Zinc. 0.0956 

Brass 0.0939 

Glass  .  .  0.1937 


Cast  iron 0. 1298 

Lead 0.0314 

Platinum 0.0324 

Silver ....  0.0570 

Tin 0.0562 

Ice 0.5040 

Sulphur 0.2026 

Charcoal.  .  .  0.2410 


PHYSICS. 

LIQUIDS. 


61 


Water  .  . . 
Alcohol . . 
Mercury  . 
Benzine . . 
Glycerine 


1.0000 
0.7000 
0.0333 
0.4500 
0.5550 


Lead  (melted) 0. 0402 

Sulphur  (melted) 0.2340 

Tin  (melted) 0.0637 

Sulphuric  acid 0.3350 

Oil  of  turpentine 0.4260 


GASES. 


Gases. 

Constant  Pressure. 

Constant  Volume. 

Air        

0  23751 

0  16847 

Oxvsren  . 

0  21751 

0  15507 

Nitrogen  

0.24380 

0  17273 

Hydrogen 

3  40900 

2  41226 

Superheated  steam  .... 
Carbonic  oxide         .  .  . 

0.48050 
0  24790 

0.34600 
0  17580 

Carbonic  acid  

0.40400 

0  15350 

111.  The  reason  that  there  are  two  values  for  the  spe- 
cific heat  of  gases  is  that  it  takes  less  heat  to  raise  the  tem- 
perature of  a  gas  when  the  volume  is  constant  than  when 
the  pressure  is  constant,  but  the  volume  varies.  Thus,  con- 
sider a  closed  cylinder  filled  with  gas.  If  heat  is  applied, 
the  pressure  and  temperature  will  both  increase.  Denoting 
the  specific  heat  at  constant  pressure  by  sp  and  at  constant 
volume  by  sv,  the  number  of  heat  units  required  to  heat  the 
gas  from  /°  to  t°  will  be  svW(T  —  T).  If,  however,  the  cyl- 
inder is  imagined  to  be  fitted  with  a  frictionless  piston,  free 
to  move  up  or  down,  and  heat  be  applied,  the  gas  will 
expand,  overcoming  a  resistance  equal  to  the  weight  of  the 
piston  plus  the  pressure  of  the  atmosphere.  Hence,  in  addi- 
tion to  the  heat  required  to  increase  the  vibratory  movement 
of  the  molecules,  heat  is  also  required  to  overcome  the  outer 
pressure,  which  remains  constant  in  this  case.  The  number 
of  heat  units  necessary  will  then  be  sp 


62  PHYSICS.  §  4 


Mixing  Two  Bodies  of  Unequal  Temperatures. 

If  a  certain  quantity  of  water  having  a  temperature  of  40°  is 
mixed  with  a  like  quantity  having  a  temperature  of  100°,  it  is 

evident  that  the  temperature  after  mixing  will  be  - 

& 

=  70°.  But  if  5  pounds  of  water  having  a  temperature  of 
40°  is  mixed  with  5  pounds  of  copper  having  a  temperature 
of  100°,  the  temperature  after  mixing  will  not  be  70°.  The 
resulting  temperature  may  be  found  by  the  following 
formula,  provided  there  is  no  change  of  the  state  in  the 
body  (ice  melting  into  water,  etc.  )  : 

^V,  +  *£*.*,+  **>.*,  +  etc. 


in  which  /  is  the  final  temperature  of  the  mixture  ;  Wl9  s^ 
and  /„  the  weight,  specific  heat,  and  temperature,  respect- 
ively, of  one  body;  l}\,  J2,  and  /2,  the  same  for  second  body; 
and  W^,  s3,  and  /3,  the  same  for  a  third  body,  etc. 

Remembering  that  the  specific  heat  of  water  is  1,  and 
getting  the  specific  heat  of  copper  from  Table  5,  the  tem- 
perature /  will  be 

5X1X40  +  5X.  0951X100 

5X1  +  5X.0951  :  4o-21 

EXAMPLE  1.  —  If  21  pounds  of  water  at  a  temperature  of  52°  are  mixed 
with  40  pounds  of  water  at  a  temperature  of  160°,  what  is  the  tempera- 
ture of  the  mixture  ? 

SOLUTION.  —  Since  the  specific  heat  of  water  is  1,  it  maybe  left  out  in 
applying  formula  27,  and  the  temperature  is  found  to  be 
21X58  +  40X160 
21+40 

EXAMPLE  2.  —  A  copper  vessel  weighing  2  pounds  is  partly  filled  with 
water  having  a  temperature  of  80°  and  weighing  7.8  pounds.  A  piece 
of  wrought  iron  weighing  3  pounds  4  ounces  and  having  a  temperature 
of  780°  is  dropped  into  this  water.  What  is  the  final  temperature  of 
the  mixture  ? 

SOLUTION.  —  Substituting  the  values  given  in  formula  27,  and  remem- 
bering that  the  original  temperatures  of  the  copper  vessel  and  the  water 
that  it  contains  are  the  same  (3  Ib.  4  oz.  =  3.25  lb.),  we  have 
_  2  X.  0951  X  80  +  7.8  X  80  +  3.25  X.  1138X780  _  , 

2X.095T+7.8  +  3.25X.H38 

Ans. 


§  4  PHYSICS.  63 

EXAMPLE  3.  —  A  wrought-iron  ball  weighing  1  pound  is  placed  in  a 
reheating  furnace  :  when  it  has  attained  the  temperature  of  the  furnace,  it 
is  taken  out  and  placed  in  a  copper  vessel  weighing  \  pound  and  contain- 
ing exactly  2  pounds  of  water  at  a  temperature  of  75".     Assuming  that 
no  water  escapes  as  steam,  and  that  the  temperature  of  the  ball,  water, 
and  vessel  after  mixing  is  156',  what  is  the  temperature  of  the  furnace  ? 
SOLUTION.  —  Substituting  the  values  given  in  formula  27, 
1  X  .1138  X  A  +  2  X  75  +  .5  X  .0951  X  75 
1  X-  1138  +  2  +  .  5  X-  0951 

.1138  /,  +153.56625 
156  =  2.16135  --  ' 

or  156  X  2.16135  =  .1138  /,  +  153.56625. 

Hence,  .1138/,  =  183.60435, 

183.60435  ,  ;,       A 

or  /,  =  —  1138      =  1,613.4  .     Ans. 

113.     By  means  of  formula  27,  the  specific  heat  of  a 
substance  may  be  obtained.      Thus,  in 

^V.+  r^V.+  rr 

^^-h/F252+/f 

suppose  that  the  specific  heat  s3  is  required  and  that  all  of 
the  other  quantities,  including"  /,  are  known. 
Then,  solving  the  above  equation  for  s,^ 


or 


EXAMPLE.  —  A  silver  vessel  weighing  13  ounces  is  suspended  by  a 
string;  1  pound  4  ounces  of  water  having  a  temperature  of  120°  is 
poured  into  it,  and  in  this  is  placed  a  piece  of  metal  weighing  14 
ounces  and  having  a  temperature  of  100°.  If  the  temperature  of'the 
vessel  was  72°-,  and  the  temperature  of  the  mixture  is  f  17°,  what  is  the 
specific  heat  of  the  piece  of  metal  ? 

SOLUTION.  —  Using  formula  28,  and  letting  IV  {,  Si,  and  A  represent, 
respectively,  the  weight,  specific  heat,  and  temperature  of  the  silver 
vessel;  \V^,  ja,  and  /2,  the  same  for  the  water;  and  ^F3,  j3,  and  /3,  the 
same  for  the  piece  of  metal, 


13X.057(72-117)  +  20X1(120-117)  -33.345  +  60       ...       . 

14(117-100)  ~238— 

All  weights  must  be  reduced  to  either  pounds  or  ounces 
before  substituting.  ' 


64  PHYSICS.  §  4 

EXAMPLES  FOR  PRACTICE. 

114.  Solve  the  following : 

1.  How  many  units  of  heat  are  required  to  raise  the  temperature 
of  10  ounces  of  platinum  from  80°  to  2,000°  ?  Ans.  38.88  B.  T.  U. 

2.  In  order  to  determine  the  specific  heat  of  a  certain  alloy,  a  piece 
weighing  12£  ounces  was  heated  to  a  temperature  of  320°,  and  was 
then  immersed  in  2  pounds  6  ounces  of  water  contained  in  a  lead  vessel 
weighing  4  pounds  7  ounces.     The  temperature  of  the  water  and  of  the 
vessel  being  70°,  what  was  the  specific  heat  of  the  alloy  if  the  tempera- 
ture of  the  mixture  was  79°  ?  Ans.  .1202. 

3.  In  order  to  determine  the  temperature  of  the  interior  of  a  chim- 
ney, a  silver  bar  weighing  20  ounces  is  placed  in  it  until  it  has  attained 
the  same  temperature.     It  is  then  immersed  in  1  pound  of  water  con- 
tained in  a  brass  vessel  weighing  10  ounces.     The  temperature  of  the 
vessel  and  water  being  65°,  and  of  the  mixture  981°,  what  was  the  tem- 
perature within  the  chimney  ?  Ans.  596°. 

4.  (a)  An  iron  casting  weighing  3  tons   is   cooled  from  2,100°  to 
100° ;  how  many  units  of  heat  does  it  give  up  ?     (b)  If  all  this  heat 
could  be  utilized,  how  many  pounds  of  coal  would  it  be  equivalent  to, 
assuming  that  1  pound  of  coal  gives  out  14,500  B.  T.  U.  during  its 
combustion  ?  .         (  (a)     1,557,600  B.  T.  U. 

S'  (  (b)    107. 4'2  Ib. 

iLATENT    HEAT. 

115.  In  all  that  has  been  said  in  the  preceding  pages 
only  the  phenomena  relating  to  sensible  heat  have  been  con- 
sidered.    If  a  quantity  of  pounded  ice  at  a  temperature  of 
32°  be  put  in  a  vessel  and  held  over  the  flame  of  a  spirit 
lamp,  heat  passes  rapidly  into  the   ice  and  melts  it;  but  a 
thermometer  resting  in  this  mixture  of  ice  and  water  shows 
no  tendency  to  rise ;  it  will  remain  at  32°  until  all  of  the  ice 
has  been  melted.     Where  has  the  heat  gone  that  was  sup- 
plied to  the  ice?     This  question  was  first  investigated  by  Dr. 
Black,  of  Edinburgh,  in  1760,  and  is  easily  explained  by  the 
modern  dynamical  theory  of  heat. 

Dr.  Black  took  a  pound  of  water  and  a  pound  of  ice,  both 
having  a  temperature  of  32°,  and  placed  them  in  two  vessels 
suspended  in  a  chamber  that  was  kept  at  as  nearly  uniform  a 
temperature  as  possible.  At  the  end  of  half  an  hour  the 
temperature  of  the  water  was  39. 2°,  but  the  ice  did  not  reach 


§  4  PHYSICS.  65 

that  temperature  till  10|  hours  had  passed,  being-  melted,  of 
course,  in  the  meantime.  Dr.  Black  reasonably  assumed 
that  the  ice  received  the  same  quantity  of  heat  that  the  water 
did  in  each  half  hour,  because  it  was  placed  in  exactly  the 
same  position  in  regard  to  the  surrounding  air;  that  is  to 
say,  it  received  39. 2  —  32  =  7.2  units  of  heat  every  half  hour, 
or  14.4  units  every  hour,  and  14.4  X 10J  =  151.2  units  in  10£ 
hours;  hence,  it  took  151.2  —  7.2  =  144  units  of  heat  to 
change  the  pound  of  ice  at  32°  into  water  at  32°.  More 
accurate  determinations  have  fixed  this  number  as  142.65, 
and  this  value  will  be  used  in  this  section  whenever  the  occa- 
sion arises  for  using  it.  If  a  pound  of  water  having  a 
temperature  of  212°  is  mixed  with  a  pound  of  water  having 
a  temperate  of  32°,  the  temperature  of  the  mixture  will  be 

21 2    I   °9 

— IL —  =  122° ;  the  boiling  water  giving  up  90°  and  the  cold 
A 

water  receiving  90°,  thus  bringing  both  to  a  common  tem- 
perature. If  a  pound  of  ice.  at  a  temperature  of  32°  is  mixed 
with  a  pound  of  water  at  a  temperature  of  212°,  the  temper- 
ature of  the  mixture  will  be  only  50.675°,  instead  of  122, 
as  in  the  previous  case.  Here  the  water  has  given  up 
212  —  50.675  =  161. 325  units  of  heat  in  order  to  bring  both 
bodies  to  a  common  temperature.  Since  the  temperature 
of  the  ice  was  raised  from  32°  to  50.675°,  it  follows  that 
50. 675  —  32  =  18. 675  units  of  heat  were  used  to  raise  the  tem- 
perature of  the  ice  after  it  had  been  melted  into  water,  and 
161.325  —  18.675  =  142.65  units  of  heat  were  necessary  to 
convert  the.  ice  at  32°  into  water  of  the  same  temperature. 
This  extra  number  of  units  of  heat,  which  is  necessary  to 
convert  a  solid  into  a  liquid  of  the  same  temperature  without 
raising  the  temperature  of  the  solid,  is  called  the  latent  'heat 
of  fusion,  and  the  temperature  at  which  this  change  of 
state  in  the  body  takes  place  is  called  the  melting  point,  or 
temperature  of  fusion.  All  solids  probably  have  a  latent 
heat  of  fusion,  the  word  probably  being  used  because  some 
solids  have  never  been  melted,  except  at  such  high  temper- 
atures that  accurate  measurements  are  not  possible,  but  its 
value  varies  greatly  for  different  substances,  being  greater 


66  PHYSICS.  §  4 

for  ice  than  for  any  other  known  solid,  while  for  frozen  mer- 
cury its  value  is  only  5. 09 ;  that  is,  to  change  1  pound  of 
frozen  merctiry  at  its  temperature  of  fusion  (—37.8°  F.)  into 
liquid  mercury  of  the  same  temperature  requires  only  5.09 
units  of  heat.  Now,  it  is  reasonable  to  suppose  that  if  it 
requires  142.65  units  of  heat  to  convert  a  pound  of  ice  at  32° 
into  water  at  32°,  then  the  same  number  of  heat  units  would 
be  given  up  when  water  at  32°  is  changed  into  ice  at  32°; 
experiment  has  shown  that  this  is  true. 

116.  If  water  is  heated  to  its  boiling  point  of  212°  under 
a  constant  pressure  of  14. 09  pounds  per  square  inch,  it  has 
been  found  by  experiment  that  it  will  require  about  966  units 
of  heat  per  pound  of  water  to  change  it  into  steam  at  212°. 
This  extra  number  of  units  of  heat  necessary  to  convert  a 
liquid  into  a  gas,  or  rather  vapor,  of  the  same  temperature 
and  pressure,  is  called  the  latent  heat  of  vaporization,  and 
the  temperature  at  which  this  change  of  state  takes  place  is 
called  the  temperature  of  vaporization. 

117.  According  to  the  modern  theory  of  heat,  the  extra 
quantity  of  heat  necessary  for  a  change  of  state  of  a  body  is 
used  in  forcing  the  molecules  of  a  body  farther  apart,  and  in 
overcoming  the  force  of  cohesion.     This  latent  heat  is  not 
lost,  but  performs  work  in  giving  additional  energy  to  the 
molecules  of  a  body,  and  it  always  reappears  when  the  body 
resumes  its  former  state.     Thus,  for  instance,  a  pound  of 
steam  under  a  pressure  of  1  atmosphere  contains  966 -|- 180 
=  1,146   units  of  heat  more  than  a  pound  of  water  at  32°. 
Hence,  if  1  pound  of  steam  at  212°  is  mixed  with  -f--fj-  =  5.37 
pounds  of  water  at  32°,  the  temperature  of  the  mixture  will 
be  exactly  212°,    or  the  boiling  point  of  water;    in   other 
words,  the  steam  raised  5.37  pounds  of  water  from  the  freez- 
ing point  to  the  boiling  point  without  lowering  its  own  tem- 
perature, by  merely  changing  from  steam  into  water.     If  a 
pound  of  water  at  a  temperature  of  32°  is  changed  into  ice  of 
the  same  temperature,  142.65  units  of  heat  will  be  given  up 
during  this  change  of  state. 


PHYSICvS. 


118.  In  the  following  table  are  given  the  temperatures 
of  fusion  and  of  vaporization,  and  the  latent  heats  of  fusion 
and  vaporization  whenever  they  have  been  determined  with 
sufficient  accuracy: 

TABLE  6. 


Substance. 

Tempera- 
ture of 
Fusion. 

Tempera- 
ture of 
Vaporiza- 
tion. 

Latent 
Heat  of 
Fusion. 

Latent 
Heat  of 
Vaporiza- 
tion. 

Water 

32° 

212° 

142  65 

966  6 

Mercury  
Sulphur  

-37.8° 
228  3° 

662° 

824° 

5.09 
13  26 

157 

Tin 

446° 

25  65 

Lead 

626° 

9  67 

Zinc  

680° 

1,900° 

50.63 

493 

Alcohol  

Unknown 

173° 

372 

Oil  of  turpentine  . 
Linseed  oil       .... 

14° 

313° 

600° 

124 

Aluminum  

1  400° 

Copper  . 

2  100° 

Cast  iron  

2  192° 

3  300° 

Wrought  iron  .... 
Steel      

2,912° 
2  520° 

5,000° 

Platinum  

3  632° 

Iridium  

4  892° 

119.  The  following  examples  will  show  the  purpose  of 
Tables  5  and  6: 

EXAMPLE  1. — How  many  units  of  heat  are  required  to  change  12 
pounds  of  ice  at  a  temperature  of  —20°  C.  into  steam  at  212°  F.  ? 

SOLUTION.— By  formula  19,  tf  —  (f  X  —  20)-f-32  =  —4°  F.  This  is 
equivalent  to  32°  +  4°  =  36°  F.  below  the  freezing  point.  In  Table  5, 
the  specific  heat  of  ice  was  given  as  .504;  hence,  it  will  take  12  X  36 
X  -504  =  217.728  B.  T.  U.  to  raise  the  temperature  of  12  pounds  of  ice 
from  —  4°  to  32°.  To  convert  this  ice  into  water  of  32°  will  require 
142.65  X  12  =  1,711.8  B.  T.  U.  To  raise  this  water  from  32°  to  a  tem- 
perature of  212°  will  require  12  X  180  =  2,160  B.  T.  U.  To  convert  it 
into  steam  of  212°  will  require  966.6  X  12  =  11,599.2  B.  T.  U.  The 
total  number  of  units  of  heat  required  will  be  217.728  +  1,711.8  +  2,160 
+  11,599.2  =  15,688.728  B.  T.  U.  Ans. 


68  PHYSICS.  §  4 

EXAMPLE  2.  —  How  many  units  of  heat  will  it  take  to  evaporate  25 
pounds  of  mercury  from  a  temperature  of  70°  ? 

SOLUTION.  —  The  temperature  of  vaporization  of  mercury  is  662°,  and 
the  specific  heat  is  .0333;  the  increase  in  temperature  from  70°  will  be 
662°  —  70°  =  592°.  The  number  of  units  of  heat  required  will  be 
25  X  592  X  -0333  =  492.84  heat  units.  The  latent  heat  of  vaporization 
is  157;  hence,  492.  84  +  25  X  157  =  4,417.  84  B.T.U  will  be  required.  Ans. 

12O.  A  solid  may  be  changed  into  a  liquid,  not  only  by 
melting  it,  but  also  by  dissolving  it,  as  salt  or  sugar  is  dis- 
solved in  water.  Since  the  particles  of  the  solid  body  must 
be  torn  asunder  in  opposition  to  the  forces  that  hold  them 
together,  it  is  reasonable  to  suppose  that  a  certain  amount 
of  heat  will  be  required  to  do  this.  That  such  is  a  fact  may 
be  easily  proved  by  any  one  having  a  thermometer.  Place 
a  thermometer  in  a  vessel  of  water  and  leave  it  there  until 
it  indicates  the  temperature  of  the  water;  then  put  in  the 
water  some  salt  or  sugar  and  stir  so  as  to  make  it  dissolve 
more  quickly,  and  it  will  be  found  that  the  mercury  has 
fallen  several  degrees.  In  fact,  if  any  solid  is  dissolved  in  a 
liquid  that  does  not  act  chemically  upon  it,  the  temperature 
of  the  mixture  will  be  lower  than  if  the  solid  did  not  dis- 
solve. It  is  this  principle  that  is  taken  advantage  of  in  the 
so  called  freezing  mixtures.  A  mixture  of  1  part  of  nitrate 
of  ammonia  and  1  part  of  water  will  reduce  the  temperature 
from  50°  to  4°,  a  fall  of  46°.  The  effects  are  still  more 
striking  when  both  bodies  are  solids,  one  of  which  is  already 
at  the  freezing  point.  Thus,  a  mixture  of  2  parts  of  snow 
or  finely  powdered  ice  and  1  part  of  common  salt  will  reduce 
the  temperature  from  50°  to  0°,  a  range  of  50°  ;  while  a  mix- 
ture of  4  parts  of  potash  and  3  parts  of  snow  or  powdered 
ice  will  lower  the  temperature  from  32°  to  —  51°,  a  range  of 
83°. 


Latent  heat  plays  an  important  part  in  every-day 
life.  It  takes  a  long  time  and  severe  cold  to  freeze  the  water 
of  a  river  to  any  depth,  even  though  the  thermometer  goes 
far  below  the  freezing  point.  This  is  because  142.65  units 
of  heat  must  be  given  up  by  every  pound  of  water,  after 
being  brought  to  the  freezing  point,  before  the  ice  can  form. 


§  4  PHYSICS.  69 

If  it  were  not  for  this,  the  rivers,  lakes,  and  other  bodies  of 
water  would  be  frozen  solid  as  soon  as  the  water  reached  the 
freezing  point  and  would  be  melted  as  soon  as  the  tempera- 
ture went  above  that  point.  In  the  spring,  all  of  the  snow 
on  the  hills  would  be  melted  during  a  warm  day,  and  great 
floods  would  be  the  consequence.  As  it  is,  142.65  units  of 
heat  must  be  supplied  to  every  pound  of  snow  at  32°  to  con- 
vert it  into  water  at  32°. 


EXAMPLES  FOB  PRACTICE. 

1.22.     Solve  the  following : 

1.  If  a  pound  of  steam  at  212°  and  7  pounds  of  ice  at  32°  are  mixed, 
what  will  be  the  resulting  temperature  ?  Ans.  50.5°. 

2.  How  many  units  of  heat  are  required  to  vaporize  10  pounds  of 
mercury  from  a  temperature  of  100°  ?  Ans.  1,757.146  B.  T.  U. 

3.  How  many  pounds  of  oil  of  turpentine  at  60°  can  be  vaporized  by 
1  pound  of  coal,  if  the  coal  gives  out  14 ,500  B.  T.U.  during  combustion? 

Ans.  62.56  Ib. 

4.  How  many  pounds  of  water  at  32°  can  be  vaporized  by  1  pound 
of  coal?  Ans.  12.646  Ib. 

5.  How  many  pounds  of  coal  are  required  to  raise  100  pounds  of 
wrought  iron  from  85°  to  its  melting  point  ?  Ans.  2.219  Ib. 


SOURCES    OF    HEAT. 

123.  Heat  is  derived  from  the  following  sources: 

1.  Physical  sources,  which  include  the  radiation  of  heat 
from  the  sun,  terrestrial  heat,  change  of  state  in  bodies,  and 
electricity. 

2.  Chemical  sources,  resulting  from  chemical  combina- 
tions, such  as  combustion,  oxidation,  etc. 

3.  Mechanical  sources,   comprising  friction,   percussion, 
and  pressure. 

124.  Physical  Sources. — 1.     The  most  intense  of  all 
of  the  sources  of  heat  is  the  sun.     The  majority  of  scientists 
are  of  the  opinion  that  all  of  the  heat  received  or  given  up 


70  PHYSICS.  §  4 

by  the  earth  has,  or  has  had,  its  source  in  the  sun.  It  would 
be  out  of  place  here  to  enter  into  this  theory  fully.  It  is  the 
amount  of  heat  radiated  from  the  sun  and  received  by  the 
earth  that  causes  the  change  of  seasons,  that  causes  the 
waters  in  the  rivers,  lakes,  and  seas  to  evaporate  and  form 
the  clouds,  to  be  again  precipitated  as  rain  or  snow.  With- 
out it,  no  living  thing,  animal  or  vegetable,  could  exist. 

2.  The   earth  possesses  a  heat  peculiar  to  itself,  called 
terrestrial  heat.     When  a  descent  is  made  below  the  surface, 
the  temperature  is  found  to  gradually  increase.     This  is  not 
caused  by  the  heat  radiated  from  the  sun,  for  the  material 
comprising  the  earth  is  such  a  poor  conductor  that  the  heat 
of  the  sun's  rays  penetrates  only  a  very  short  distance  below 
the  surface.     The  explanation  usually  given  for  this  phe- 
nomenon is  that  the  interior  of  the  earth  is  in  a  molten  con- 
dition.    The  terrestrial,  heat  exerts  but  a  slight  effect,  not 
raising  the  temperature  of  the  surface  more  than  ^V  °f  a 
degree. 

3.  If  a  liquid  is  poured  iipon  a  finely  divided  solid,  as  a 
sponge,   flour,   starch,   roots,  etc.,   the  temperature   will  be 
increased  from  1°  to  10°,  according  to  conditions.     This  phe- 
nomenon may  be  called  heat  produced  by  capillarity. 

4.  The  heat  produced  by  a  change  of  state  has  already 
been  described ;  it  is  the  heat  given  off  when  a  body  is  con- 
verted from  a  gas  or  liquid  to  a  liquid  or  solid. 

5.  Extremely  high  temperatures   may  be  produced  by 
the  electric  current.      By  means  of  it,  quicklime,  firebrick, 
osmium,  porcelain,  and  several  other  substances,  which  until 
very  recently  have  resisted  every  attempt  to  melt  them,  may 
be  made  to  run  like  water. 

125.  Chemical  Sources. — Whenever  two  or  more  sub- 
stances that  act  chemically  upon  one  another  are  brought 
together  and  allowed  to  combine,  heat  is  evolved ;  and  when- 
ever the  heat  caused  by  this  chemical  union  is  sufficiently 
intense  to  raise  the  resultant  substances  to  a  temperature 
at  which  they  emit  light,  this  act  of  union  is  termed  com- 
bustion. 


PHYSICS. 


71 


This  subject  will  not  be  treated  here,  but  will  be  considered 
fully  by  itself  in  Theoretical  Chemistry. 


Mechanical  Sources.  —  1.  The  friction  between 
any  two  bodies  rubbed  together  produces  heat.  Rubbing 
one  hand  briskly  against  the  other  will  soon  make  the  hands 
too  warm  for  comfort.  The  friction  between  a  journal  and 
its  bearing  causes  heat;  the  heat  causes  the  journal  and 
bearing  to  expand,  the  journal  expanding  more  rapidly  on 
account  of  being  smaller  and  being  heated  more  quickly; 
the  expansion  causes  a  greater  pressure  on  the  bearing,  pro- 
ducing more  friction  and  heat.  If  the 
bearing  is  not  properly  oiled,  the  heat  will 
become  so  intense  in  a  short  time  that  the 
soft  metal  in  it  will  melt.  When  shooting 
stars  strike  the  earth's  atmosphere,  their 
velocity  is  so  great  (sometimes  as  high  as 
150  miles  a  second)  that  the  friction  of 
the  atmosphere  causes  them  to  take  fire 
almost  instantly.  Wherever  there  is  fric- 
tion, there  is  heat. 

2.  Heat  is  also  generated  by  percussion. 
The  repeated  blows  of  a  hammer  upon  a 
piece  of  iron,  lead,  or  other  metal  will  soon 
make  it  quite  hot. 

3.  The  generation  of  heat  by  pressure 
was  spoken  of  in  connection  with  gases; 
that  is,  the  .temperature  rises  when  a  gas 
is  compressed.     This  is  also  true  of  solids 
and  liquids,  but  the   results   are    not    so 
marked  in  their  cases.     The  production  of 
heat  by  the  compression  of  gases  is  easily 
shown  by  means  of  the  pneumatic  syringe 
shown  in  Fig.  17.     This  consists  of  a  glass 

tube  with  thick  sides,  hermetically  closed  with  a  leather 
piston.  At  the  bottom  is  a  small  cavity  in  which  a  piece 
of  cotton,  moistened  with  ether  or  carbon  disulphide,  is 
placed.  The  tube  being  filled  with  air,  the  piston  is  suddenly 


72  PHYSICS.  §  4 

plunged  downwards.  Thus  compressed,  the  air  generates 
so  much  heat  that  the  cotton  is  ignited,  and  can  be  seen 
burning  when  the  piston  is  suddenly  withdrawn.  The 
ignition  of  the  cotton  in  this  experiment  indicates  *a  tem- 
perature of  at  least  570°,  since  it  will  not  ignite  at  a  lower 
temperature. 


MOHT. 


PROPAGATION    OF    LIGHT. 

127.  When  the  nerves  of  the  retina  of  the  eye  are 
stimulated,  the  sensation  is  commonly  described  by  saying 
that  light  is  seen.  Light  is,  then,  a  pure  sensation;  and  it 
may  be  caused  in  several  ways.  Ordinarily  it  is  produced 
as  the  result  of  some  action  that  is  taking  place  at  some  dis- 
tance from*  the  observer  —  such  as  the  combustion  of  gas  in  a 
lamp  flame,  or  the  chemical  processes  of  the  sun. 


128.  Theory  of  Jjight.  —  Sir  Isaac  Newton  assumed  that 
luminous  bodies  emit  infinitely  small  particles  in  straight 
lines,  which,  by  penetrating  the  transparent  parts  of  the  eye 
and  falling  upon  the  nervous  tissue,  produce  light.     Modern 
philosophers,  however,  in  developing  the  theory  of  sound, 
came  to  the  conclusion  that  light  might  be  the  effect  of 
undulations,  or  little  waves,  propagated  with  inconceivable 
velocity  through   the  highly  elastic   medium   of  excessive 
tenuity  called  ether,  which  fills  all  space  and  lies  between 
the  particles  of  material  substances. 

129.  Transparent,  Translucent,  and  Opaque  Bodies. 

Bodies  that  allow  light  to  pass,  so  that  objects  can  be  clearly 
distinguished  through  them,  are  called  transparent;  the  air, 
water,  and  clear  glass  are  common  examples.  Oiled  paper, 
roughened  glass,  and  porcelain  allow  light  to  pass,  but  objects 
cannot  be  clearly  distinguished  through  them  ;  they  are  trans- 
lucent bodies.  Opaque  bodies,  such  as  wood  or  stone,  do  not 


§  4  PHYSICS.  73 

allow  light  to  pass  through  them.  Opaque  bodies,  when 
very  thin,  are  translucent,  and  transparent  bodies,  when 
very  thick,  become  translucent  and  even  opaque. 

130.  Propagation  of  right.—  A  medium   is  a  sub- 
stance, as  air  and  glass,  through  which  light  passes.     If  the 
density  and  composition  of  the  medium  is  the  same  in  all 
parts,  it  is  called   homogeneous  ;  water,  carefully  prepared 
glass,  and  the  atmosphere,  if  we  consider  only  a  small  por- 
tion of  it,  are  homogeneous  mediums.     In  a  homogeneous 
medium  light  is  propagated  in  straight  lines  ;    this  is  our 
every-day  experience.      The  rays  of  the  sun,  as  they  trace 
their  way  through  our  room,  are  straight  ;  in  order  that  light 
may  pass  through  a  tube,  the  tube  must  be  straight. 

131.  Ray  and  Pencil  of  Ijight.  —  A  light  proceeding 
from  a  luminous  body  is  supposed  to  be  made  up  of  straight 
lines,  called  rays  of  light.     A  number  of  rays  form  a  pencil 
of  light. 


Velocity  of  light.  —  A  ray  of  light  emitted  from 
a  luminous  body  proceeds  in  a  straight  line  with  extreme 
velocity.  Through  a  series  of  astronomical  observations,  an 
approximate  knowledge  of  this  velocity  has  been  obtained. 
As  the  moon  revolves  about  the  earth,  so  revolve  the  satel- 
lites of  Jupiter  about  that  planet,  and  the  time  of  revolution 
of  each  satellite  has  been  ascertained  from  its  periodical 
entry  into  or  exit  from  the  shadow  of  the  planet.  The  time 
required  by  -one  satellite  is  only  42  hours.  The  Danish 
astronomer  Romer  observed  that  this  period  appeared  longer 
when  the  earth  in  its  passage  around  the  sun  increased  the 
distance  between  Jupiter  and  itself,  and  again  that  the 
periodic  time  became  shorter  when  the  earth  moved  towards 
Jupiter.  Although  the  difference  is  rather  small  for  a  single 
revolution  of  the  satellite,  it  increases  by  the  addition  of 
many  revolutions  during  the  passage  of  the  earth  from  its 
nearest  to  its  greatest  distance  from  Jupiter,  a  passage  that 
occupies  about  6  months,  till  it  amounts  to  16  minutes  and 
16  seconds.  From  this,  Romer  concluded  that  the  light  of 


7.4 


PHYSICS. 


the  sun  reflected  from  the  satellite  required  that  time  to  pass 
through  a  distance  equal  to  the  diameter  of  the  orbit  of  -the 
earth,  and,  since  this  space  is,  approximately,  190,000,000 
miles,  the  velocity  of  light  must  be  about  190,000  miles  per 
second. 


REFLECTION    AND    REFRACTION. 

133.  Let  it  be  assumed  that  a  ray  of  light  is  deflected 
by  a  mirror  J/,  Fig.  18,  so  that  it  strikes  point  A^  of  a  hori- 
zontal mirror  L,  and  is  from  there  reflected  in  the  direction 


A*.  If  we  then  draw  a  line  from  point  N  to  O  at  right  angles 
to  the  mirror  and  measure  the  angles  M N  O  and  O  N  R,  we 
find  that  these  two  angles  are  equal,  and  we  further  notice 
that  the  two  rays  lie  in  a  plane  which  is  at  right  angles  to 
the  mirror  N.  The  line  A"  O  perpendicular  to  the  mirror  is 
called  the  normal.  The  ray  M  N  is  called  the  incident  ray 
and  the  ray  N  R  is  the  reflected  ray.  The  angle  that  the 
incident  ray  makes  with  the  normal  is  the  angle  of  incidence, 
and  the  angle  made  by  the  normal  and  the  reflected  ray  is 
the  angle  of  reflection. 

134.     The  general  laws  of  reflection  are  as  follows: 

1.  The  angle  of  incidence  is  equal  to  the  angle  of  reflec- 
tion. 

2.  In  whatever  position  the  incident  ray  meets  the  reflect- 
ing surface,  a  plane  can  always  be  so  placed  that  it  passes 
through  the  incident  ray>  'the  reflected  ray,  and  the  normal. 


§  4  PHYSICS.    -  75 

135.  The  same  laws  also  hold  good  if  a  mirror  or  any 
other  reflecting  surface  be  curved,  as  a  portion  of  a  sphere, 
since  a  curve  may  be  considered  as  made  up  of  a  multitude  of 
minute  planes.  Parallel  rays,  however,  cease  to  be  parallel 
when  reflected  from  such  a  surface ;  they  become  conver- 
gent or  divergent  according  as  the  reflecting  surface  is  con- 
cave or  convex. 

Suppose  that  A  B,  Fig.  19,  is  a  section  of  a  concave  mirror. 
Point  C,  the  center  of  the  sphere,  is  known  as  the  center  of 


FIG.  19. 

curvature,  point  O  as  the  apex,  and  CO  as  the  principal 
axis.  A  very  small  portion  of  the  mirror  at  any  point  A  will 
be  a  plane,  of  which  A  C,  the  radius,  will  be  the  normal; 
and  if  any  incident  ray,  as  S A,  strikes  the  mirror  at  A,  the 
reflected  ray  will  be  A  F  (angle  SA  C  =  angle  FA  C)\  the 
same  is  .true  for  any  small  portion  of  the  mirror,  as,  for 
instance,  L,  O,  and  B.  If  the  arc  A  B,  the  section  of  the 
reflecting  surface,  is  small  compared  to  the  radius  CO,  it 
is  found,  both  by  measurements  and  experiments,  that  all 
incident  rays,  as  SA,  TL,  X B,  parallel  to  the  principal  axis, 
after  reflection,  pass  through  the  same  point  Fin  the  prin- 
cipal axis.  This  point,  which  is  midway  between  C  and  the 
mirror,  is  called  \}^Q  principal  focus,  and  the  distance  between 
F  and  O  is  known  as  the  focal  length. 

136.     In  Art.  13O  it  was  stated  that  the  rays  of  light 
proceed  in  straight  lines;  this,  however,  is  true  only  as  long 


76 


PHYSICS. 


as  the  medium  through  which  these  rays  pass  is  homogene- 
ous. Should,  however,  the  density  or  chemical  nature  of  the 
medium  through  which  the  ray  travels  vary,  the  ray  of  light 
is  bent  from  its  original  course  into  a  new  one ;  or  it  is  said 

to  be  refracted.  Fig.  20  shows 
a  thick  plate  of  glass  through 
which  a  ray  of  light  is  sup- 
posed to  travel,  B  being  the 
point  of  the  surface  where 
the  ray  A  enters.  This  ray, 
instead  of  proceeding  onwards 
in  the  direction  C,  will  be  bent 
downwards  towards  point  D, 
and  on  leaving  the  plate  and 
entering  into  the  air  on  the  other  side,  will  again  be  bent, 
but  in  an  opposite  direction,  so  as  to  travel  parallel  to  the 
continuation  of  the  original  path,  provided  there  is  the  same 
medium  on  the  upper  as  on  the  lower  side  of  the  plate.  The 
general  law  of  refraction  may,  then,  be  expressed  as  follows: 
When  a  ray  of  light  passes  from  a  rarer  to  a  denser  medium 
it  is  usually  refracted  towards  a  line  perpendicular  to  the 
surface  of  the  latter,  and,  conversely,  when  it  leaves  a  dense, 
medium  for  a  rarer  one,  it  is  refracted  away  from  a  line  per- 
pendicular to  the  surface  of  the  denser  substance.  In  the 
former  case  the  angle  of  incidence  is  greater  than  that  of 
refraction;  in  the  latter  it  is  less.  In  Fig.  20,  angle  ABE 
is  the  angle  of  incidence,  and  D  B  E'  is  the  angle  of  refrac- 
tion. 

137.  The  amount  of  refraction,  for  the  same  medium, 
varies  with  the  obliquity  with  which  the  ray  strikes  the  sur- 
face.    When  perpendicular  to  the  latter,  the  ray  passes  with- 
out change  of  direction ;  and  in  other  positions  the  refraction 
increases  with  the  obliquity. 

138.  Let  A,  Fig.  21,  represent  a  ray  of  light  falling  upon 
the  surface  of  a  heavy  plate  glass  at  the  point  B.     From  this 
point  let  a  perpendicular  fall  and  be  continued  into  the  new 
medium,  and  around  B  as  a  center  let  a  circle  be  drawn.   Then, 


PHYSICS. 


77 


according  to  the  general  law  of  refraction,  stated  in  Art.  136, 
the  refraction  must  be  towards  the  perpendicular,  in  the 
direction  B  C,  for  example.  If  now  the  lines  a  a  and  a'  a', 
which  are  the  sines  of  the  angles  of  incidence  and  refraction, 
respectively,  are  drawn  at  right  angles  to  the  perpendiculars, 
and  their  lengths  compared  by  means  of  a  scale,  these  lengths 
will  be  found,  in  this  case,  to  be  in- the  ratio  of  3  :  2.  Or,  if 
another  ray  such  as  E 
is  taken,  which  is  A 

refracted  in  the  same 
manner  towards  Fy  the 
bending  being  natu- 
rally greater  from  the 
increased  obliquity  of 
the  ray,  and  if  the 
sines  of  the  new  angles 
of  incidence  and  re- 
fraction are  then  com- 
pared, they  will  be 
found  to  bear  to  each 
other  the  same  ratio;  i 
expressed  by  the  rule  : 


FIG.  21. 


e.,  3  :  2.     This  fact  maybe,  then, 
So  long  as  the  light  passes  from  one 

to  the  other  of  tivo  media,  the  ratio  of  the  sines  of  the  angles 

of  incidence  and  refraction  remains  constant. 


139.  Different  bodies  possess  different  refractive  power, 
and,  as  a  rule,  the  densest  substances  refract  most.  The 
method  adopted  for  describing  the  comparative  refractive 
power  of  different  bodies  is  to  state  the  ratio  of  the  sine  of 
the  angle  of  incidence  in  the  first  medium  to  the  sine  of  the 
angle  of  refraction  in  the  second  medium ;  this  ratio  is  known 
as  the  index  of  refraction  of  the  two  substances.  It  is  greater 
or  less  than  unity,  according  as  the  second  medium  is  denser 
or  rarer  than  the  first. 

If  the  index  of  refraction  of  any  particular  substance  is 
once  known,  the  effect  of  the  latter  upon  a  ray  of  light  enter- 
ing it  at  any  angle  can  be  easily  calculated  by  the  law  of 
sines. 


PHYSICS. 


14O.  In  Table  7  the  indices  of  refraction  of  several 
substances  are  given.  It  is  assumed  that  the  ray  of  light 
passes  into  the  substances  in  question  from  the  air. 

TABKE    7. 

INDICES  OF  REFRACTION. 


Ice 1.30 

Water 1.34 

Fluorspar 1.40 

Plate  glass 1.50 

Rock  crystal 1.60 

Chrysolite 1.69 


Carbon  bisulphide 1.70 

Garnet 1.80 

Phosphorus 2. 20 

Diamond 2.50 

Lead  chromate 3.00 

Cinnabar  .  .  3.20 


141.  Some  Results  of  Refraction. — A  stick  partly  sub- 
merged in  water  appears  to  be  bent  at  the  surface ;  the  tip  of 
the  stick  and  all  that  is  under  water  seems  to  be  raised.     The 
atmosphere  is  less  dense  the  higher  we  ascend,  and  we  can 

assume  it  to  be  made  up 
of  layers  of  air  of  various 
densities.  The  rays  of 
light  from  the  sun  or  a 
star,  instead  of  descending 
in  straight  lines,  are  re- 
fracted at  every  layer. 
Thus,  the  light  from  St  in. 
Fig.  22,  reaches  the  eye  as 
if  it  came  from  Sf ;  that  is, 
a  star  appears  higher  in 
FIG-  **•  the  heavens  than  it  really 

is.     For  the  same  reason,  the  sun  is  seen  before  it  is  above, 

and  after  it  is  below,  the  horizon. 

142.  Irregular   Refraction. — The    rays  of   light    are 
straight  in  air  and  straight  in  water,  provided  the  density  is 
uniform.     This  is  not  the  case  if  from  any  cause  the  density 
is  not  uniform.     The  quivering  of  objects,  when  seen  over 
heated  coke  or  on  a  hot  day,  is  due  to  unequal  refraction, 
the  density  being  constantly  changing. 


PHYSICS. 


79 


1 43.  Total  Reflection. — When  a  ray  of  light  passes  from 
one  medium  into  another  that  is  less  refracting,  as  from  water 
into  air,  it  has  been  seen  in  Art.  136  that  the  emerging  ray 
is  again  bent  from  its  course.   Hence,  when  light  is  propagated 
in  a  mass  of  water  from  CtoA,  Fig.  23,  there  is  always  a  value 
of  the  angle  of  incidence  CAB 

such  that  the  angle  of  refrac- 
tion fiADisa  right  angle,  in 
which  case,  the  refracted  ray 
emerges  parallel  to  the  surface 
of  the  water.  This  angle  CAB 
is  called  the  critical  angle, 
since,  for  any  greater — E  A  B, 
for  instance — the  incident  ray 
can  no  longer  emerge,  but 
undergoes  reflection.  This  is 
called  total  reflection,  because 
the  incident  light  is  entirely  reflected.  The  critical  angle 
for  water  and  air  is  48^-°. 

If  you  raise  a  glass  of  water  above  your  head  you  can 
easily  find  a  position  in  which  the  rays  from  the  eye  to  every 
part  of  the  under  surface  make  a  greater  angle  than  48^° 
with  the  normal.  A  brilliant  mirror  due  to  the  total  reflec- 
tion of  light  will  be  observed. 

144.  Transparent   glass   when  finely  ground  loses  its 
transparency;    light    is    diffused    at   the    surface,    and    the 
few   rays   that   pass   through   the  upper  layer  of   the  fine 
particles  are  refracted  in  passing  from   the   glass   to    the 
air    between   the    surfaces;    the   result   of    this    combined 
irregular  reflection  and  refraction  is  that  the  glass  appears 
opaque. 

If  a  liquid,  whose  refractive  index  is  the  same  as  that  of 
glass,  is  poured  over  the  glass,  the  result  will  be  the  same, 
in  respect  to  the  action  of  light,  as  if  the  surface  were  made 
smooth  again  and  all  the  spaces  were  filled  with  glass;  that 
is,  powdered  glass  becomes  in  its  effects  a  glass  plate  and.  is 
again  transparent. 


80  PHYSICS.  §  4 

145.  Prisms. — A  prism  may  be  defined  as  a  portion  of 
a  transparent  medium  bounded  by  plane  inclined  surfaces. 
The  line  along  which  these  surfaces  meet  is  called  the 
refracting  edge,  and  the  angle  between  them,  the  refracting 
angle  of  the  prism.  The  side  opposite  the  refracting  edge 
is  called  the  base  of  the  prism,  and  any  section  through  it, 
perpendicular  to  this  edge,  is  called  a  principal  section.  If 
a  ray  of  light  falls  upon  one  of  the  inclined  surfaces  of  a 
prism,  the  change  in  the  direction  of  the  ray — that  is,  its 
deviation — instead  of  being  neutralized  at  the  second  surface, 
is  still  further  increased. 

Let  ray  A,  Fig.  24,  be  incident  to  the  surface  PQof  a 
prism  at  O*  At  O  the  refraction  will  be  towards  the  normal 


•N 


FIG.  24. 

and  the  ray  will  be  refracted  to  O'.  Draw  normals  at  O 
and  O'.  At  O'  the  second  refraction  takes  place,  but  since 
the  face  PS  is  inclined  to  the  face  P Q,  the  second  deviation 
is  in  the  same  direction  as  the  first;  that  is,  towards  the  base 
of  the  prism.  Evidently  the  deviation  of  the  ray  at  the  first 
surface  is  A  ON—  t  O  O'  or  d  =  i  —  r,  and  at  the  second  is 
A'O'N'—tO'Oora"  =  i'-r'.  The  total  deviation  is  the  sum 
of  the  partial  deviations,  or  D  —  d-\-df  =  (i—  r)-\-(t'  —  r'). 
That  is,  the  total  deviation  in  the  figure,  or  the  angle  A  I />, 
or  D,  between  the  incident  and  emergent  rays  is  equal  to  the 
sum  of  the  partial  deviations  d-\-d',  or  I O  O'  and  I  O'  O, 
since  the  exterior  angle  of  the  triangle  is  equal  to  the  sum 
of  the  two  interior  angles.  Moreover,  the  deviation  may  be 


PHYSICS. 


81 


obtained  in  terms  of  the  index  (;/)  and  the  prism  angle. 
Since  D  —  (i—r)-\-  (i'  —  r')  or  (i  +  i')  —  (r  +  r'),  and  as  the 
prism  angle  is  equal  to  r-\-rf,  the  deviation  D  =  (i-^-i')  —a. 
If  the  prism  be  thin  and  the  incident  ray  nearly  normal,  we 
may  write  /  =  nr  and  i'  =  nr',  whence  i-\-i'  =  n(r-\-rr) 
or  //  <7,  and  D  —  n  a  —  a  or  (//  —  1)  a\  which  means  that  the 
deviation  is  equal  to  the  prism  angle  multiplied  by  the 
index  of  refraction  less  1. 


146.  When  the  deviation  produced  by  a  prism  is  a 
minimum,  the  angles  formed  by  the  ray  with  the  normals 
within  the  prism  are  equal,  as  are  also  the  angles  without  it; 
that  is,  r  =  r'  and  /  =  i'. 
Consequently,  we  can 
simplify  the  formula 
for  deviation,  D 

writing  it  D  =  2z  — 2r. 

In  Fig.  25  we  see  that 

the  angle  at  t  formed 

by  the  two  normals  is 

for  that  reason  equal  FIG.  25. 

to  #,  the  refractive  angle  of  the  prism.     But,  being  exterior 

to  the  triangle   O  O' t,  it  is  equal  to  the  sum  of  the  two 

interior  angles,  that  is,  7'-(-r/or2r;   therefore,  a  =  2r  and 

r  —  \a.       Hence,  D  —  2/  — <?;    and  i  =  ±(D-\-a).     By  the 


formula  for  the  index,  n  = 


sn 


we  have  n  = 


sn 


a) 


sin  r  sin  \a 

To  determine  the  refractive  index  of  a  substance,  there- 
fore, it  is  only  necessary  to  measure  (1)  the  angle  a  .of.  the 
prism  and  (2)  the  deviation  angle  D  when  this  angle  has  its 
minimum  value. 


147.  Lenses. — A  lens  is  any  portion  of  a  transparent 
medium  bounded  by  curved  surfaces,  and  may  be  considered 
as  being  made  up  of  a  large  number  of  prisms.  Suppose  we 
have  a  number  of  small  prisms  and  we  place  the  prism  By 


PHYSICS. 


Pig.  26,  with  the  largest  angle  so  that  a  ray  of  light  A  is 
refracted  to  C\    then,  evidently,  if  a  prism  //,  exactly  like 


G 


B,  be  placed  as  in  the  figure,  it  also  will  refract  light  to  C. 

By  using  D  and  G  with  smaller 
angles  (less  deviation),  we  can 
find,  by  trial  or  calculation,  posi- 
tions at  which  they  also  refract 
rays  from  A  to  C\  we  similarly 
use  E  and  F,  and  between  E  and 
F  we  might  place  a  plate  of 
glass  with  parallel  sides.  Thus, 
a  number  of  rays  from  A  can  be 
made  to  converge  to  C.  The 
arrangement  shown  at  (b),  Fig. 
27,  will  have  exactry  the  same 
effect  as  (a),  Fig.  27,  if  the  sur- 
faces of  A,  B,  C,  etc.  in  (b)  are  inclined  at  the  same  angles 
as  the  surfaces  of  A,  B,  C,  etc.  in  (a).  A  very  large  num- 
ber of  such  prisms  fitted  together  as  in  (^),  Fig.  27,  will 
form  a  body  whose  section  is  as  represented  at  (c),  and  the 
surface  will  be  practically  spherical. 

It  has  been  found  by  experiments  and  calculations  that 
lenses  whose  surfaces  are  parts  of  the  surface  of  a  sphere  act 
like  the  prisms  in  («),  Fig.  27,  if  the  surfaces  are  small 
compared  with  the  whole  surface  of  the  sphere. 

148.  The  names  given  to  the  lenses  in  Fig.  28  are:  A, 
double  convex — both  surfaces  convex ;  B,  plano-convex — one 
surface  convex,  one  plane;  C,  concavo-convex  converging — 


PHYSICS. 


83 


one  surface   convex,  one  concave;  D,  double  concave — both 
surfaces  concave ;  E,  plano-concave — one  concave,  one  plane ; 


FIG.  28. 

Ft  concavo-convex  diverging — one  concave,  one  convex.     E 
and  F  are  also  known  as  meniscus  lenses. 

149.  Surfaces  of  lenses  are  usually  parts  of  the  surfaces 
of  spheres  of  equal  radii.  Fig.  29  shows  how  a  double- 
convex  lens  is  formed.  The  line  A  B  passing  through  the 


FIG.  29. 

center  is  the  principal  axis;  C n  and  Cm  are  the  normals  at 
n  and  /;/,  and  C,  C'  are  the  centers  of  curvature.  The  point 
o  is  called  the  optical  center.  The  point  where  the"  rays 
parallel  to  the  principal  axis  come  to  a  focus,  after  refrac- 
tion, is  called  the  principal  focus.  (See  Art.  135.) 


THE  SPECTRUM. 

15O.  Dispersion. — It  is  an  every-day  experience  in 
experiments  with  prisms  and  lenses  to  observe  that  the 
images  are  tinged  with  colors,  reminding  us  of  the  hues  of 
the  rainbow.  These  colors  are  due  to  refraction.  If  a  ray  of 
light  is  admitted  into  a  dark  room  through  a  small  hole  in 
a  shutter,  or  otherwise,  and  is  allowed  to  fall  upon  a  glass 


84  PHYSICS.  §  4 

prism,  and  if  a  screen  is  put  in  the  patli  of  the  rays,  as 
shown  in  Fig.  30,  there  is  refraction;  but,  instead  of  an 
image,  we  find  on  the  screen  a  band  of  color,  called  a  spec- 
trum. This  separation  of  light  into  its  component  parts  is 
called  dispersion.  The  light  in  passing  through  the  prism 
has  been  decomposed  and  we  can  distinguish  in  order  red 
(the  least  refracted  of  the  rays),  orange,  yellow,  green,  blue, 
indigo,  and  violet  (the  most  refracted). 


FIG.  30. 


Sir  Isaac  Newton  was  the  first  to  investigate  this  phe- 
nomenon, and  drew  from  it  the  inference  that  white  light  is 
composed  of  these  seven  primitive  colors,  the  ravs  of  which 
are  differently  refrangible  by  the  same  medium  and  are  thus 
capable  of  being  separated. 

151.  Bodies  of  the  same  refractive  power,  of  course,  do 
not  equally  disperse  the  differently  colored  rays  to  the  same 
extent;  if,  for  instance,  the  principal  yellow  or  red  rays  are 
equally  refracted  by  two  prisms  of  different  substances,  it 


§  4  PHYvSICS.  85 

does  not  necessarily  follow  that  the  blue  or  violet  rays  are 
also  similarly  affected.  Hence,  prisms  of  different  varieties 
of  glass  or  other  transparent  substances  give,  under  similar 
circumstances,  very  different  spectra,  both  in  regard  to  the 
length  of  the  image  and  the  relative  extent  of  the  colored 
band. 


FIG.  si. 

152.  As  the  appearance  of  the  spectrum  also  varies  with 
the  nature  of  the  light,  an  apparatus  called  a  spectroscope 
has  been  invented  to  investigate  these  variations.     Fig.  31 
illustrates  the  principle  of  this  instrument.     The  apparatus 
is  so  arranged  that  the  light  after  passing  through  a  fine  slit 
^  impinges  upon  a  flint-glass  prism  P,  by  which  it  is  dis- 
persed.    The  decomposed  light  emerges  from  the  prism  in 
several  directions  between  R  (red  rays)  and  V  (violet  rays) ; 
and  the  spectrum  thus  produced  is  observed  by  the  telescope 
T.     Only  a  part  of  the  spectrum  enters  the  telescope  at  one 
time,  but  the  several  parts  may  be  readily  examined   by 
sliglitly  turning  either  the  telescope  or  the  prism. 

153.  If  sunlight  is  allowed  to  fall  upon  the  slit  of  the 
spectroscope,  numerous  fine  black  lines,  of  different  degrees 
of  breadth  and  shade,  are  observed.     These  lines  are  always 
present,  and  always  occupy  exactly  the  same  relative  position 
in  the  solar  spectrum.     They  indicate  the  absence  in  sunlight 
of  particular  rays,  and  they  may  be  considered  as  shadows  or 
spaces  where  there  is  no  light.     They  were  discovered  in 
1802  by  Dr.  Wollaston,  and  subsequently  were  more  thor- 
oughly  investigated   by   Fraunhofer,    a   German    optician. 
They  are  generally  known  as  Fraunhofer 's  lines.     Some  of 
them,  in  consequence  of  their  peculiar  strength  and  relative 


86  PHYSICS.  §  4 

position,  may  always  be  easily  recognized;    the  more  con- 
spicuous are  shown  in  Fig1.  32. 


FIG.  32. 

154.  In  the  last  few  years  the  existence  of  these  lines 
has  become  a  matter  of  very  great  importance,  as  through 
them  the  determination  of  the  chemical  constitution  of  the 
sun  and  fixed  stars  has  become  possible.     The  spectra  of  the 
moon  and  planets,    which  are  indirectly  produced  by  the 
reflected  light  of  the  sun,  exhibit,  of  course,  the  same  lines 
in  unaltered  position,  while  in  the  spectra  of  the  fixed  stars, 
though  dark  lines  also  occur,  these  stellar  lines  are  entirely 
different  from    those  seen  in  direct  or  reflected   sunlight. 
Hence,  the  conclusion  has  been  drawn  that  the  Fraunhofer's 
lines  are  in  some  way  produced  in  the  body  of  the  sun  itself, 
but  it  is  only  recently  that  the  cause  of  their  production  has 
been  discovered   by   Kirchhof   and    Bunsen,    and    thus   the 
foundation  laid  for  the  science  of  solar  and  stellar  chemistry. 

155.  It  has  long  been  known  that  certain  chemical  sub- 
stances, especially  the  salts  of  the  alkalies  and  alkaline  earths, 
when  strongly  heated  in  the  blowpipe  flame,  or  other  nearly 
colorless  flames,  impart  to  that  flame  a  peculiar  color,  by  the 
occurrence  of  which  the  presence  of  these  substances  can  be 
detected.     If  these  colored  flames  are  examined  by  means  of 
a  spectroscope,  it  is  at  once  seen  that  the  light  thus  refracted 
differs  essentially  from  white  light,  because  it  consists  only 
of  a  particular  set  of  rays,  each  flame  giving  a  spectrum  con- 
taining a  few  bright  bands.     The  spectrum  of  the  yellow 
soda  flame,  for  instance,  contains  only  one  fine  bright  yellow 
line,  while  the  purple  potash  flame  exhibits  a  spectrum  in 
which  two  bright  lines  appear,  one  lying  at  the  extreme  red 


§  4  PHYSICS.  87 

end  and  the  other  at  the  extreme  violet  end.  These  peculiar 
lines  are  always  produced  by  the  same  chemical  element,  and 
by  no  other  known  substance,  and  the  position  of  these  lines 
always  remains  unaltered.  When  the  spectrum  of  a  flame 
tinted  by  a  mixture  of  sodium  and  potassium  salts  is  exam- 
ined, the  yellow  ray  of  sodium  is  found  to  be  confined  to  its 
own  position,  while  the  potassium  red  and  purple  lines  are 
just  as  plainly  visible  as  they  would  have  been  had  no 
sodium  been  present. 

Each  of  the  colored  flames  exhibited  by  the  salts  of  stron- 
tium, barium,  lithium,  and  calcium  likewise  produces  a 
peculiar  spectrum.  By  means  of  these  spectra  the  presence 
or  absence  of  very  minute  quantities  of  these  substances 
when  mixed  together  can  be  ascertained  by  simply  observing 
the  presence  or  absence  of  the  peculiar  bright  bands  charac- 
teristic of  the  particular  body. 

156.  If  the  positions  of  the  dark  lines  in  the  solar  spec- 
trum are  carefully  compared  in  a  powerful  spectroscope  with 
those  of  the  bright  lines  in  the  spectra  of  certain  metals, 
such,  for  instance,  as  iron,  magnesium,  and  sodium,  it  is  seen 
that  each  of  the  bright  lines  of  the  particular  metal  coincides 
not  only  in  position,  but  also  in  width,  with  a  dark  solar  line ; 
so  that  if  a  spectroscope  is  so  arranged  that  a  solar  and 
metallic  spectrum  is  suffered  to  fall  one  below  the  other  in  the 
field  of  the  telescope,  the  bright  lines  of  the  metal  all  appear 
as  continuations  of  the  dark  solar  lines.  In  the  case  of  iron 
alone  more  than  sixty  such  coincidences  have  been  observed. 
With  other  metals,  such  as  gold,  antimony,  lithium,  etc. ,  no 
single  coincidence  has  been  noticed,  while  all  the  lines  of 
certain  other  metals  have  their  dark  representatives  in  the 
sun's  spectra. 

These  coincidences  could  not  be  a  mere  chance.  They 
gave  rise,  therefore,  to  the  hypothesis  that  certain  dark  lines 
in  the  spectra  of  the  sun,  which  are  coincident  with  the 
bright  lines  in  the  spectra  of  iron,  indicate  the  presence  of 
iron  in  the  sun.  This  hypothesis  was  strengthened  through 
experiments  of  Foucault  in  1849,  who  had  reversed  the  yellow 


88  PHYSICS.  §  4 

double  line  in  the  spectrum  of  the  electric  arc  into  a  double 
dark  one.  Kirchhof  further  investigated  this  subject  and 
found  that  in  like  manner  the  spectra  of  many  substances 
could  be  reversed,;  thus,  the  bright  yellow  soda  lines,  coin- 
cident with  Fraunhofe'r's  line  D  (see  Fig.  32),  can  be  made 
to  appear  as  a  dark  line  by  allowing  the  rays  of  a  strong 
white  line  to  pass  through  a  flame  colored  by  soda,  and  then 
to  fall  upon  the  slit  of  a  spectroscope.  Instead  of  then  seeing 
the  usual  soda  spectrum,  consisting  of  a  bright  yellow  double 
line  upon  a  dark  ground,  a  double  dark  line  identical  in  posi- 
tion and  width  with  the  soda  line  will  appear  to  intersect  the 
spectrum  of  the  white  light. 

This  fact,  being  the  missing  connecting  link,  at  once 
thoroughly  established  the  theory  that  by  observing  the 
coincidences  of  these  dark  lines  with  the  bright  lines  of  our 
metals,  we  are  able  to  arrive  at  a  knowledge  of  the  occur- 
rence of  such  metals  in  the  solar  atmosphere. 

157.  The  attention  of  the  student  may  here  be  drawn 
to  the  fact  that  light  and  heat  are  reflected  and  refracted 
similarly,  mirrors  and  lenses  having  practically  the  same 
focus  for  each.  Between  the  smallest  particles  of  matter 
there  is  supposed  to  exist  a  medium  called  ether;  it  is  by  the 
vibration  of  the  ether  spheres  that  light  and  radiant  heat  are 
transmitted.  The  waves  are  of  different  lengths ;  that  is,  the 
periods  of  vibration  of  the  particles  differ.  The  longest  rays 
cause  the  sensation  of  heat,  next  come  the  rays  that  cause 
the  sensation  of  red  color,  and  shorter  than  these  are  the 
rays  that  cause  the  sensation  of  orange  color.  As  the  rays 
become  shorter  the  other  colors  of  the  spectrum  are  pro- 
duced, the  shortest  of  the  light  waves  giving  the  sensation 
of  violet  color.  There  are  yet  still  shorter  rays  than  the 
violet,  known  as  actinic  rays  ;  they  do  not  affect  the  eye  as 
regards  color,  but  are,  however,  important  in  photography, 
from  their  power  to  affect  certain  chemical  substances. 

When  a  body  is  heated  it  possesses  energy;  this  energy 
sets  the  ether  spheres  in  motion,  and  the  first  waves  trans- 
mitted are  the  heat  waves.  As  the  temperature  of  the  body 


§  4  PHYSICS.  89 

rises,  the  increased  energy  starts  other  waves.  First  we  have 
those  that  cause  the  sensation  of  red  color;  the  body  is  at  red 
heat.  As  the  temperature  further  rises  shorter  waves  are 
started,  and  the  color  changes  until,  when  all  the  light  waves 
are  in  motion,  the  composite  effect  is  white  light;  the  body 
is  at  white  heat,  as  we  commonly  express  it. 

Ice  transmits  the  light  waves,  but  absorbs  the  heat  waves, 
whose  energy  melts  the  ice.  A  solution  of  iodine  in  carbon 
disulphide  absorbs  the  light  waves — that  is,  no  light  passes — 
but  transmits  the  heat  waves.  Red  glass  allows  the  long 
waves  to  pass,  but  absorbs  the  short  waves. 

A  simple  color,  then,  depends  upon  the  wave  length,  that 
is,  upon  the  period  of  vibration;  the  intensity  of  such  a  color 
depends  upon  the  amplitude  of  the  vibrations.  The  wave 
lengths  are,  naturally,  exceedingly  small,  there  being  33,000 
waves  per  inch  in  a.  red  ray,  and  64, 000  per  inch  in  a  violet 
ray. 

DOUBMiJ  REFRACTION  AXL>  POLARIZATION. 

158.  Double  Refraction. — A  large  number  of  crystals 
possess  the  property  of  double  refraction,  which  means  that 
if  a  ray  of  light  passes  through  such  a  crystal,  it  becomes 
split  or  divided  into  two  rays;  one,  which  is  called  the  ordi- 
nary ray,  follows  the  general  law  of  refraction,  while  the 
other,  which  is  known  as  the  extraordinary  ray,  takes  an 
entirely  different,  new  course,  which  depends  on  the  position 
of  the  crystal.  This  remarkable  property  is  found  in  nearly 
all  bodies  that  crystallize  in  any  other  than  the  cubical  form. 
The  phenomenon  is  explained  by  assuming  that  the  ether  in 
these  doubly  refracting  bodies  is  not  equally  elastic  in  all 
directions,  and  that,  consequently,  the  vibrations,  in  certain 
directions  at  right  angles  to  each  other,  are  transmitted  with 
unequal  velocity.  A  transparent  variety  of.calcite — found 
in  Iceland,  and  hence  known  as  Iceland  spar — possesses  this 
property  in  a  marked  degree.  If  a  piece  of  this  substance  is 
placed  upon  a  sheet  of  white  paper  on  which  an  ink  spot  or 
any  other  mark  has  previously  been  made,  two  images  of  the 
mark  will  be  seen. 


90  PHYSICS.  §  4 

159.  Polarization. — If  a  ray  of  light  CD  is  thrown 
upon  a  plate  of  glass  A  A,  as  shown  in  Fig.  33,  at  an  angle 
of  56°  45',  that  portion  of  the  ray  which  is  reflected  will  be 
found  to  have  acquired  properties  it  did  not  before  possess ; 
for,  on  throwing  it  at  the  same  angle  upon  a  second  glass 
plate  B  B,  it  will  be  observed  that  there  are  two  particular 
positions  of  the  latter  (indicated  by  Bf  B') ;  namely,  those  in 


which  the  planes  of  incidence  are  at  right  angles  to  each 
other,  when  the  ray  of  light  is  no  longer  reflected,  but 
entirely  refracted.  Light  that  has  undergone  this  change  is 
said  to  be  polarized. 

The  other  portion  of  the  ray  that  passes  through  the  first, 
or  polarizing,  plate  A  A  is  also  to  a  certain  extent  in  this 
peculiar  condition — i.e.,  polarized — and  by  using  a  number  of 
similar  plates,  placing  one  behind  the  other,  this  effect  may 
be  greatly  increased.  It  must,  however,  be  remarked  that 
the  light  polarized  by  transmission  is  in  an  opposite  state  to 
that  polarized  by  reflection;  that  is,  when  examined  by 
the  second,  or  analyzing,  plate,  held  at  the  angle  before 
mentioned,  it  will  be  seen  to  be  reflected  when  the  other  is 
transmitted,  and  to  be  dispersed  when  the  first  is  reflected. 

16O.  Angle  of  Polarization. — Not  every  transparent 
substance  is  capable  of  polarizing  light.  Glass,  water,  and  a 
number  of  other  substances,  however,  produce  this  change, 


§  4  PHYSICS.  91 

each  having  a  particular  angle  at  which  the  effect  is  the 
greatest.  The  angle  of  polarization  of  a  substance  is,  then, 
the  angle  that  the  incident  ray  must  make  with  the  perpen- 
dicular to  a  plane  surface  of  that  substance  in  order  that  the 
polarization  be  complete.  For  glass  this  angle  is  56°  45';  for 
water,  52°  45';  for  quartz,  57°  32';  for  diamond,  68°;  etc. 
.  If  in  the  preceding  experiment  the  plane  had  been  inclined  at 
a  larger  or  smaller  angle,  the  ray  of  light  would  not  have  been 
completely  polarized,  which  would  be  shown  by  the  ray  being 
partially  reflected  from  the  analyzing  plate  B  B  in  all  positions. 

161.  Polarization  in  Opposite  Directions. — The  ordi- 
nary and  extraordinary  rays  produced  by  sending  a  ray  of 
light  through  a  substance  possessing  the  property  of  double 
refraction  are  found  on  examination  to  be  completely  polar- 
ized, but  in  a  very  peculiar  manner.     The  one  is  capable  of 
reflection  when  the  other  is  transmitted  or  refracted.     This 
fact  is  expressed  by  saying  that  the  two  rays  are  polarized  in 
opposite  directions.     With  a  good  sized  piece  of  transparent 
Iceland  spar,  the  two  oppositely  polarized  rays  maybe  widely 
separated  and  examined  separately  by  an  analyzing  plate. 

162.  Some  of  the  doubly  refracting  substances  absorb 
one  of  these  rays,  but  not  the  other.    Tourmaline,  a  mineral, 
is  the  best  known  medium  possessing  this  property.     When 
a  ray  of  light  is  sent  through  a  plate  of  tourmaline  cut  at 
right  angles  to  the  axis,  an  ordinary  and  an  extraordinary 
ray  are  formed,  which  are  polarized  in  planes  at  right  angles 
to  each  other.     The  mineral,  however,  rapidly  absorbs  the 
ordinary  ray,  and  if  the  emer- 
ging polarized  extraordinary  ray 

is  observed  through  a  second 
similar  plate,  held  exactly  in  the 
same  position  as  the  first  one 
and  as  shown  in  Fig.  34  (a), 
little  change  will  be  noticed; 
but  when  the  second  plate  is 
turned  around  as  shown  in  Fig. 
34  (#),  so  as  to  make  the  axes  cross  at  right  angles,  it  will  be 


92  PHYSICS.  §  4 

found  that  the  light  is  almost  wholly  stopped.  A  plate  of 
this  mineral  in  this  way  is  a  valuable  test  for  detecting 
polarized  light,  and  light  that  has  not  undergone  the  change. 
Instead  of  tourmaline  plates,  which  are,  owing  to  the  thick- 
ness required  for  complete  polarization,  not  very  transpar- 
ent, and  are  mostly  more  or  less  colored,  two  Nicol's  prisms, 
or  the  cheaper  and  quite  as  effective  Foucault's  prisms,  are 
used,  as  they  possess  the  same  properties  as  tourmaline. 

Two  of  these  prisms,  placed  in  exactly  the  same  position, 
one  behind  the  other,  allow  light  to  pass  through  them ;  but 
by  turning  the  second  prism  a  cloudiness  is  produced,  until, 
by  continuing  to  turn  the  prism,  a  point  is  reached  that 
causes  perfect  darkness.  This  phenomenon  occurs  whether 
the  light  is  colored  or  not. 

163.  INTicol's  prism  consists  of  a  bar  of  Iceland  spar, 
about  1  inch  in  height  and  |  inch  in  breadth ;  the  prism  is 
cut  in  the  plane  that  passes  through  the  obtuse  angles,  as 
shown  in  Fig.  35  (a).  The  two  halves  are  joined  by  a  layer 


(*> 

FIG.  35. 

of  Canada  balsam.  The  principal  involved  in  the  construc- 
tion of  this  prism  is  the  following:  The  refractive  index  of 
Canada  balsam,  being  1. 549,  is  less  than  the  index  of  the  ordi- 
nary ray  of  Iceland  spar,  1. 654,  and  greater  than  the  index 
of  its  extraordinary  ray,  1.483;  hence,  if  a  ray  of  light  CD, 
Fig.  35  (/;),  enters,  the  prism,  the  ordinary  ray  is  totally 
reflected  on  the  surface  A  B  and  takes  the  direction  D  E  F, 
by  which  it  is  refracted  out  of  the  prism,  while  the  extraor- 
dinary ray  D  G  emerges  alone.  The  prism  is  used  as  an 
analyzer  and  also  as  a  polarizer,  possessing  the  same  prop- 
erties as  tourmaline,  but,  as  has  been  already  mentioned,  is 
preferable  because  it  gives  a  colorless  field  of  view. 


§  4  PHYSICS.  93 

In  Foucatilt's  prism,  which  is  constructed  on  exactly  the 
same  principle,  the  layer  of  Canada  balsam  is  replaced  by  a 
layer  of  air,  the  prism  being  held  together  by  a  brass  or 
aluminum  mounting. 

164.  Circular  Polarization. — A  ray  of  light  colored 
by  passing,  for  example,  through  a  plate  of  red  glass,  which 
has  passed  through  the  first  of  two  Nicol's  prisms,  and  which 
is  then  obstructed  in  consequence  of  the   position   of   the 
second  prism,  regains  its  original  property  of  penetrating 
the  second  prism  if  a  plate  of  rock  crystal  is  placed  between 
the  two  prisms  in  the  path  of  the  ray.     The  passage  of  the 
ray  through  the  analyzer  (second  prism)   can  now  only  be 
interrupted  by  a  further  turn  of  the  analyzer.     The  rotation 
required  depends  on  the  thickness  of  the  plate  and  the  color 
of  the  ray.     It  increases  from  red  in  the  following  order: 
yellow,  green,  blue,  and  violet. 

This  kind  of  polarization  is  known  as  circular  polarization, 
and  for  a  long  time  quartz  or  rock  crystal  was  the  only  solid 
substance  known  to  exhibit  this  peculiar  property  of  produ- 
cing it.  Scientific  investigations  and  experiments,  however, 
have  brought  to  light  since  then  a  number  of  substances  that 
possess  this  property  in  a  still  higher  degree;  thus,  for 
instance,  a  plate  of  cinnabar  acts  nearly  fifteen  times  more 
powerfully  than  a  plate  of  equal  thickness  of  rock  crystal. 

The  direction  of  the  rotation  that  is  required  to  produce 
this  complete  opaqueness  of  the  analyzer  depends  on  the 
composition  of  the  interposed  substance.  A  substance  is  said 
to  be  dcxtro-  or  levo-rotary,  according  as  it  is  necessary  to 
turn  the  analyzer  around  in  the  direction  of  the  hands  of  a 
watch,  or  in  the  opposite  direction.  The  angle  through  which 
the  polarizer  has  to  be  turned  is  called  the  angle  of  rotation. 

165.  It  has  been  observed  that  solutions  of  many  organic 
substances  show  the  property  of  circular  polarization,  thoiigh, 
as  a  rule,  to  a  less  extent  than  rock  crystal.     Thus,  solutions 
of  cane  sugar  and  tartaric  acid  possess  right-handed  polariza- 
tion, while  oil  of  turpentine  and  albumin  are  left-handed. 
In  all  of  these  solutions  the  angle  of  rotation  varies  directly 


94  PHYSICS.  §  4 

with  the  strength  of  the  solutions,  and  the  thickness  of  the 
column  through  which  the  light  passes;  hence,  circular 
polarization  has  become  an  important  auxiliary  in  chemical 
analysis. 

This  power  of  circular  polarization  is  found  most  com- 
monly in  vegetable  and  animal  products  and  their  immediate 
derivatives,  as  will  be  seen  in  the  description  of  such  bodies 
in  Organic  Chemistry.  It  is  assumed  that  this  power  depends 
on  some  peculiarity  in  the  structure  of  the  molecules. 

166.  In  order  to  determine  the   amount  of  polarization 
any  liquid  may  exhibit,  it  is  poured  into  a  glass  tube  and 
closed  with  glass  plates.     This  is  then  placed  between  two 
Nicol's  prisms,  which  have  previously  been  so  arranged  that 
the  analyzer  appears  opaque.    Through  the  polarizing  power 
of  the  liquid,  the  analyzer  becomes  transparent,  and  from 
the  magnitude  of  the  angle  through  which  the  analyzer  has 
to  be  turned  to  appear  opaque  again,  the  polarizing  power  of 
the  liquid  is  calculated. 

167.  An  apparatus  of  this  kind  is  widely  used  and  is  of 
the  greatest  importance  in  the  sugar  industry.     It  is  called 
the  saccliarimeter,  and  is  used  for  determining  the  con- 
centration of  solutions  of  cane  sugar.     The  instrument  was 
devised  by  Biot,  and  through  certain  additions  perfected  by 
Soleil. 

Though  a  full  description  of  the  saccharimeter,  its  uses, 
and  instructions  in  regard  to  calculations,  etc.  cannot  be 
given  in  this  section,  a  short  explanation  of  this  apparatus 
will  enable  the  student  to  comprehend  more  fully  the  impor- 
tance of  circular  polarization  in  connection  with  chemistry, 
and  .will  give  him  some  idea  as  to  the  way  in  which  this 
physical  phenomenon  is  titilized. 

This  apparatus  is  shown  in  Fig.  3G.  The  two  Nicol's 
prisms  are  in  the  'fastenings  A  and  /?,  respectively.  After 
the  prisms  are  arranged  so  that  the  analyzer  A  appears 
opaque,  the  tube,  filled  with  a  solution  of  sugar,  is  placed  as 
marked  by  dotted  lines  in  the  figure.  The  light  will  now 


PHYSICS. 


95 


be  again  enabled  to  pass  through  the  analyzer,  and  prism  A 
has  to  be  turned  through  a  certain  angle  till  entire  darkness 
again  appears.  The  magnitude  of  this  angle  is  read  on  the 
circular  disk  C,  which  is  divided  into  degrees,  and  upon 
which  moves  an  index  D  connected  with  the  prism  A. 

If  we  assume  that  the  tube  used  is  exactly  10  inches  long 
and  the  solution  to  be  analyzed  contains  exactly  10  per  cent. 


FIG.  36. 

of  cane  sugar,  the  angle  of  rotation  will  be  19.6°.  Now,  the 
magnitude  of  this  angle  is  directly  proportional  to  the  length 
of  the  liquid  column  and  also  to  the  percentage  of  sugar  in 
Solution.  If,  therefore,  a  solution  containing  x  per  cent,  of 
sugar  in  a  tube  y  inches  long  produces  an  angle  of  rotation 
of  z  degrees,  the  percentage  of  sugar  will  be  given  by  the 
equation  : 


y     x 
":  ioxio' 


or?  *    = 


100-s- 


(29.) 


The  apparatus  shown  in  Fig.  36  is  of  the  simplest  kind, 
and  can  only  be  used  conveniently  when  pure  cane  sugar  is 
to  be  analyzed;  if  the  solution  contains  cane  sugar  and 
uncrystallizable  sugar,  the  latter  possessing  left-handed 


96  PHYSICS.  §  4 

rotation  and  the  former  right-handed,  only  the  difference  of 
these  two  actions  is  obtained.  The  whole  quantity  has  to  be 
changed  into  uncrystallizable  sugar,  the  experiment  repeated, 
and  from  the  results  of  the  two  observations  the  quantity  of 
each  kind  of  sugar  calculated.  Soleil  has  added  to  this  form 
of  saccharimeter  a  number  of  improvements  that  facilitate 
observations  and  make  measurements  more  exact. 


MAGNETISM. 

168.  Magnets  are  substances  that  Have  the  property  of 
attracting  pieces  of  iron  or  steel,  and  the  term  magnetism  is 
applied  to  the  cause  of  this  attraction.  Magnetism  exists  in 
a  natural  state  in  an  ore  of  iron,  which  is  known  in  chemistry 
as  magnetic  oxide  of  iron  or  magnetite.  This  magnetic  ore 
was  first  found  by  the  ancients  in  Magnesia,  a  city  in  Asia 
Minor;  hence,  the  name  magnet. 

It  was  also  discovered  that  when  a  small  bar  of  this  ore  is 
suspended  in  a  horizontal  position  by  a  thread,  it  has  the 
property  of  pointing  in  a  north-and- south  direction.  From 
this  fact  the  name  lode  stone,  "leading  stone,"  was  given  to 
the  ore. 

When  a  bar  or  needle  of  hardened  steel  is  rubbed  with  a 
piece  of  lodestone,  it  acquires  magnetic  properties  similar  to 
those  of  the  lodestone,  without  the  latter  losing 
any  of  its  own  force.      Such  bars  are   called 
artificial  magnets. 

Artificial  magnets  that  retain  their  magnet- 
ism for  a  long  time  are  called  permanent  mag- 
nets. 

The  common  form  of  artificial  or  permanent 
magnet,  Fig.  37,  is  a  bar  of  steel  bent  into  the 
shape  of   a  horseshoe  and  then  hardened  and 
magnetized.      A  piece  of  soft  iron  called  an 
armature,  or  a  keeper,  is  placed  across  the  two 
FIG.  37.         free  ends  and  helps  to    prevent    the  magnet 
from  losing  its  magnetic  force. 


PHYSICS. 


1 69.  If  a  bar  magnet  is  dipped  into  iron  filings,  the  filings 
are  attracted  towards  the  two  ends  and  adhere  there  in  tufts ; 
while  towards  the  center  of  the  bar,  half-way  between  the 
two  ends,  there  is  no  such  tendency  (see  Fig.  38).  That  part 
of  the  magnet  where  there  is  no  apparent  magnetic  attrac- 


FlG.  38. 

tion  is  called  the  neutral  line ;  and  the  parts  around  the 
ends  where  the  attraction  is  greatest  are  called  poles.  An 
imaginary  line  drawn  through  the  center  of  the  magnet  from 
end  to  end,  connecting  the  two  poles  together,  is  the  axis 
of  'magnetism. 

17O.  A  compass  consists  of  a  magnetized  steel  needle, 
Fig.  39,  resting  upon  a  fine  point  and  capable  of  turning 
freely  in  a  horizontal  plane.  When  not 
in  the  vicinity  of  other  magnets  or  mag- 
netized iron,  the  needle  will  always  come 
to  rest  with  one  end  pointing  towards  the 
north  and  the  other  towards  the  south. 
The  end  pointing  northward  is  the  north- 
seeking  pole,  or,  simply,  the  north  pole, 
and  the  opposite  end  is  the  south- sec  king 
pole,  or  south  pole.  This  polarity  applies  as  well  to  all  mag- 
nets. 

If  the  north  pole  of  one  magnet  is  brought  near  the  south 
pole  of  another  magnet,  attraction  takes  place;  but  if  two 
north  poles  or  two  south  poles  are  brought  together,  they 
repel  each  other.  In  general,  like  magnetic  poles  repel  eacJi 
other;  unlike  poles  attract  each  other. 


98 


PHYSICS. 


171.  The  earth  is  a  great  magnet  whose  magnetic  poles 
coincide  nearly,  but  not  quite,  with  the  true  geographical 
north  and  south  poles.  A  freely  suspended  magnet,  there- 
fore, will  always  point  in  a  north-and-south  direction. 

173.  It  is  impossible  to  produce  a  magnet  with  only 
one  pole.  If  a  long  bar  magnet  is  broken  into  any  number 
of  parts,  each  part  will  still  be  a  magnet  and  have  two  poles 
— one  north  and  one  south  pole. 

173.  Substances  that  are  not  in  themselves  magnets — 
that  is,  do   not   possess   poles   and   neutral   lines,   but   are 
capable  of  being  attracted  by  a  magnet — are  called  magnetic 
substances.     In  addition  to  iron  and  its  alloys,  the  following 
metals  are  magnetic  substances:   nickel,  cobalt,  manganese, 
cerium,   and   chromium.      These    metals,   however,   possess 
magnetic  properties  in  a  very  inferior  degree  compared  with 
iron  and  its  alloys.     All  other  known  substances  are  called 
non-magnetic  substances. 

174.  The  space  surrounding  a  magnet  is  called  a  mag- 
netic field;  or,  in  other  words,  a  magnetic  field  is  a  place 
where  a  freely  suspended  magnetic  needle  will  always  come 
to  rest,  pointing  in  the  same  direction. 


^\\^)\\;Y,U!  1 1  \Hfur'/'W''£'% 


FIG.  40. 


FIG.  41. 


175.  Magnetic  attractions  and  repulsions  are  assumed 
to  act  in  a  definite  direction  and  along  imaginary  lines  called 
lines  of  magnetic  force,  or,  simply,  lines  of  force,  and  every 
magnetic  field  is  assumed  to  be  traversed  by  such  lines  of 


PHYSICS. 


99 


force.  Their  position  in  any  plane  may  be  shown  by  placing 
a  sheet  of  paper  over  a  magnet  and  sprinkling  fine  iron  filings 
over  the  paper.  In  the  case  of  a  bar  magnet  lying  on  its 
side,  the  iron  filings  will  arrange  themselves  in  curved  lines 
extending  from  the  north"  to  the  south  pole,  as  shown  in 
Fig.  40.  A  view  of.  the  magnetic  field  looking  towards  either 
pole  of  a  bar  magnet  would  exhibit  merely  radial  lines,  as 
shown  by  the  filings  in  Fig.  41. 


FIG.  42. 


Every  line  of  force  is  assumed  to  pass  out  from  the  north 
pole,  make  a  complete  circuit  through  the  surrounding 
medium,  and  into  the  south  pole,  thence  passing  through  the 
magnet  to  the  north  pole  again,  as  shown  in  Fig.  42.  This 


NOTE. — In  all  diagrams,  the  direction  of  the  lines  of  force  will  be 
represented  by  arrowheads  upon  dotted  lines. 


100 


PHYSICS. 


is  called  the  direction  of  tJic  lines  of  force,  and  the  path  they 
take  is  called  the  magnetic  circuit. 

The  direction  of  the  lines  of  force  in  any  magnetic  field 
can  be  traced  by  a  small,  freely  suspended  magnetic  needle 
or  a  small  compass,  such  as  indicated  by  m  in  Fig.  42.  The 
north  pole  of  the  needle  will  always  point  in  the  direction  of 
the  line  of  force,  the  length  of  the  needle  lying  tangent  to 
the  lines  of  force  at  that  point.  If  the  needle  is  moved 
bodily  in  the  direction  towards  which  the  north  pole  points, 
its  center  or  pivot  will  describe  a  path  coinciding  with  the 
direction  of  the  lines  of  force  along  that  part  of  the  magnetic 
field. 

Lines  of  force  can  never  intersect ;  when  two  opposing  mag- 
netic fields  are  brought  together,  as  indicated  by  the  iron 
filings  in  Figs.  43  and  44,  the  lines  of  force  from  each  will 


ymm^ 

•y/im 


FIG.  43. 


FIG.  44. 


be  crowded  and  distorted  from  their  original  direction  until 
they  coincide  in  direction  with'  those  opposing  and  form  a 
resultant  field  in  which  the  direction  of  the  lines  of  force  will 
depend  on  the  relative  strengths  of  the  two  opposing  nega- 
tive fields.  The  resulting  poles  thus  formed  are  called  con- 
sequent poles. 

In  every  magnetic  field  there  are  certain  stresses  that 
produce  a  tension  along  the  lines  of  force  and  a  pressure 
across  them  ;  that  is,  they  tend  to  shorten  themselves  from 
end  to  end,  and  repel  one  another  as  they  lie  side  by  side. 


§  4  PHYSICS.  101 

17(>.  When  a  magnetic  substance  is  brought  into  a  mag- 
netic field,  the  lines  of  force  in  that  vicinity  crowd  together 
and  all  tend  to  pass  through  the  substance.  If  the  substance 
is  free  to  move  on  an  axis,  but  not  bodily  towards  the  mag- 
net pole,  it  will  always  come  to  rest  with  its  greatest  extent 
or  length  in  the  direction  of  the  lines  of  force.  The  body 
will  then  become  a  temporary  magnet,  its  south  pole  being 
situated  where  the  lines  of  force  enter  it,  and  its  north  pole 
where  they  pass  out.  The  production  of  magnetism  in  a 
magnetic  substance  in  this  manner  is  called  magnetic  induc- 
tion. The  production  of  artificial  magnetism  in  a  hardened- 
steel  needle  or  bar  by  contact  with  lodestone  is  only  a  special 
case  of  magnetic  induction. 

177.  The  amount,  or  quantity,  of  magnetism  is  expressed 
by  the  total  number  of  lines  of  force  contained  in  a  magnetic 
circuit. 

178.  Magnetic  density  is  the  number  of  lines  of  force 
passing  through   a  unit  area  measured  perpendicularly  to 
the  direction  of  the  lines. 


ELECTRICITY. 


INTRODUCTION. 

179.  Electricity  is  the  name  given  to  that  which 
directly  causes  all  electrical  phenomena.  The  word  is 
derived  from  the  Greek  word  vJAe/crpov,  meaning  "amber." 

Although  electrical  science  has  made  great  advances  in  the 
last  few  years,  the  exact  nature  of  electricity  is  unknown. 
Recent  researches  tend  to  demonstrate  that  all  electrical 
phenomena  are  due  to  a  peculiar  state  or  stress  of  a  medium 
called  ether ;  that,  when  in  this  condition,  the  *et1ier  pos- 
sesses potential  energy,  or  capacity  for  doing  ^vork,  as  is 


102  PHYSICS.  §  4 

manifested  by  attractions  and  repulsions,  by  chemical  decom- 
position, and  by  luminous  heating,  and  various  other  effects. 

180.  In  all  probability,  electricity  is  not  a  form  of  mat- 
ter, for  the  only  physical  properties  it  possesses  in  common 
with  material  substances  are  indestructibility  and  elasticity  ; 
it  does  not  possess  'cv  eight,  extension,  nor  any  of  the  other 
physical  properties  of  matter. 

181.  Electrical  science  is  founded  upon  the  effects  pro- 
duced by  the  action  of  certain  forces  upon  matter,  and  all 
knowledge  of  the  science  is  deduced  from  these  effects.     The 
study    of  the  fundamental  principles  of  the  science  is  an 
analysis  of  a  series  of  experiments  and  the  classification  of 
the  results,  under  laws  and  rules.     It  is  not  necessary  to 
keep  in  mind  any  hypothesis  as  to  the  exact  nature  of  elec- 
tricity; its  effects  and  the  laws  that  govern  them  are  quite 
similar   to    those    of   well   known    mechanical    and    natural 
phenomena,  and  will  be  best  understood  by  comparison. 


Electricity  may  either  appear  to  reside  upon  the 
surfaces  of  bodies  as  a  charge,  under  high  pressure,  or  flow 
through  their  substance  as  a  current,  under  comparatively 
low  pressure.  That  branch  of  science  which  treats  of  charges 
upon  the  surfaces  of  bodies  is  termed  electrostatics,  and  the 
charges  are  said  to  be  static  charges. 

Electrodynamics  is  that  branch  which  treats  of  the  action 
of  electric  currents. 


ELECTROSTATICS. 

183.  When  a  glass  rod  or  a  piece  of  amber  is  rubbed 
with  a  piece  of  silk  or  fur,  the  parts  rubbed  will  be  found  to 
have  the  property  of  attracting  light  bodies,  such  as  pieces 
of  silk,  wool,  feathers,  gold  leaf,  pith,  etc.,  which,  after 
momentary  contact,  are  repelled.  These  attractions  and 
repulsions  are  caused  by  a  static  charge  of  electicity,  residing 
upon  the  surfaces  of  those  bodies.  A  body  in  this  condition 
is  said  to  be  electrified. 


PHYSICS. 


103 


FIG.  45. 


A  better  experiment  for  demonstrating  this  action  is  to 
suspend  a  small  pith  ball  by  a  silk  thread  from  a  support  or 
bracket,  as  shown  in  Fig.  45.  Such 
an  apparatus  is  spoken  of  as  an 
electric  pendulum.  If  a  static  charge 
of  electricity  is  developed  on  a  glass 
rod  by  rubbing  it  with  silk,  and  the 
rod  brought  near  the  pendulum, 
the  ball  will  be  attracted  to  the  rod ; 
but,  after  momentary  contact,  it 
will  be  repelled.  By  this  contact 
the  ball  becomes  electrified,  and, 
so  long  as  the  two  bodies  retain 
their  charges,  mutual  repulsion  will 
take  place  whenever  they  are 
brought  near  each  other.  If  a  stick  of  sealing  wax,  electrified 
by  being  rubbed  with  fur,  is  brought  near  to  another  pendu- 
lum, the  same  results  will  be  produced — the  ball  will  fly 
towards  the  wax,  and,  after  contact,  will  again  be  repelled. 
But  the  charges  respectively  developed  in  the  preceding  cases 
are  not  in  the  same  condition.  For  if,  after  the  pith  ball  has 
been  touched  with  the  glass  rod  and  repelled,  the  electrified 
sealing  wax  is  brought  in  the  vicinity,  attraction  takes 
place  between  the  ball  and  the  sealing  wax.  Similarly,  if  the 
pendulum  is  charged  with  the  electrified  sealing  wax,  the 
ball  will  be  repelled  by  the  wax  and  attracted  by  the  glass  rod. 

An  electric  charge  excited  upon  glass  by  rubbing  it  with 
silk  has  been .  termed  a  positive  charge  (+) ;  and  that  devel- 
oped on  resinous  bodies  by  friction  with  flannel  or  fur,  a 
negative  charge  (—). 

184:.  Neither  charge  is  produced  alone,  for  there  is 
always  an  equal  quantity  of  both  charges  produced,  one 
charge  appearing  on  the  body  rubbed,  and  an  equal  amount 
of  the  opposite  charge  upon  the  person  or  thing  rubbing. 

185.  The  intensity  of  the  charge  developed  by  rubbing 
the  two  substances  together  is  independent  of  the  actual 


104  PHYSICS.  §  4 

amount  of  friction  that  takes  place  between  the  bodies.  For, 
in  order  to  obtain  the  highest  possible  degree  of  electrifica- 
tion, it  is  only  necessary  to  bring  every  portion  of  one  surface 
into  intimate  contact  with  every  particle  or  every  portion  of 
the  other  surface;  when  this  is  done,  no  extra  amount  of 
rubbing  can  develop  any  greater  charge  upon  either  sub- 
stance. 

186.  From  these  experiments  are  derived  the  following 
laws  : 

1.  When  two  dissimilar  substances  are  placed  in  contact, 
one  of  them  ahvays  assumes  the  positive  and  the  other  the 
negative  condition,  altJiough  the  amount  may  sometimes  be  so 
small  as  to  render  its  detection  very  difficult. 

2.  Electrified  bodies  with  similar  charges  are  mutually 
repellent,  ivliile  electrified  bodies  with  dissimilar  charges  are 
mutually  attractive. 

187.  In  the  following  list,  called  the  electric  series,  the 
substances  are  arranged  in  such  order  that  each  receives  a 
positive  charge  when  rubbed  with  any  of  the  bodies  follow- 
ing, and  a  negative  charge  when  rubbed  with  any  of  those 
preceding  it. 

1.  Fur  G.  Cotton  11.  Sealing  wax 

2.  Flannel  7.  Silk  12.  Resin 

3.  Ivory  8.  The  body  13.  Sulphur 

4.  Crystals  9.  Wood  14.  Gutta  percha 

5.  Glass  10.  Metals  15.  Guncotton 

For  example,  glass,  when  rubbed  with  fur,  receives  a  neg- 
ative charge ;  but  when  rubbed  with  silk  it  receives  a  positive 
charge. 

188.  An  electroscope  is  an  instrument  for  detecting 
static  charges  of  electricity,   and  for  determining  whether 
they   are   positive    or   negative ;    it   does   not  measure   the 
intensity  of  the  charges. 

The  pith  ball  suspended  by  a  silk  thread  acts  as  a  simple  elec- 


PHYSICS. 


105 


troscope.  A  more  sensitive  electroscope  is  showrrin  Fig.  40, 
and  consists  of  two  gold 
leaves  suspended  with- 
in a  glass  jar  J,  which 
serves  to  protect  them 
from  drafts  of  air  and  to 
support  them  from  con- 
tact with  the  earth.  The 
gold  leaves  a  are  sup- 
ported side  by  side  in 
the  jar  by  a  brass  rod 
or  wire  b,  which  passes 
through  a  cork  in  the 
mouth  of  the  jar.  The 
upper  end  of  the  brass 
rod  is  furnished  with  a 
flat  metallic  plate  or  ball  c.  An  electrified  body,  such  as  the 
rod  d,  brought  into  the  vicinity  of  the  electroscope  will  cause 
the  leaves  to  repel  each  other,  and  they  will  remain  in  that 
position  for  some  time. 

To  determine  tJie  condition  of  a  charge  by  tlie  electroscope: 
First,  charge  the  gold  leaves  with  a  known  charge,  such  as 
that  developed  upon  glass  when  rubbed  with  silk.  The 
leaves  will  spread  apart,  being  electrified  with  a  positive 
charge.  When  they  are  thus  charged,  the  approach  of  a 
"body  positively  charged  will  cause  them 
to  open  still  more  widely;  while  on  the 
approach  of  one  negatively  charged, 
they  will  close  together. 


FIG.  46. 


189.       The  torsion  balance  is  an 

instrument  used  to  measure  the  force 
exerted  between  two  electrified  bodies. 
It  consists  of  an  arm  or  lever  of  some 
light  insulating  material,  such  as  a  straw 
or  piece  of  wood,  provided  at  one  end 
with  a  gilt  pith  ball  ?/,  Fig.  47,  and  sus- 
pended in  a  glass  jar  by  a  fine  silver  wire.     The  wire  passes 


FIG.  47 


106  PHYSICS.  §  4 

up  through  a  glass  tube  and  is  fastened  to  a  brass  stopper  b, 
called  the  torsion  Jiead.  'The  torsion  head  is  graduated  in 
degrees  and  is  capable  of  being  revolved  around  upon  the 
glass  tube.  Another  gilt  pith  ball  m  is  fastened  to  the  end 
of  the  vertical  glass  rod  a,  which  is  inserted  through  an  open- 
ing in  the  top  of  the  jar.  A  narrow  strip  of  paper,  also 
divided  into  degrees,  encircles  the  glass  jar  at  the  level  of 
the  two  pith  balls. 

To  use  the  torsion  balance :  Turn  the  torsion  head  around 
until  the  two  pith  balls  m  and  n  just  touch  each  other. 
Remove  the  glass  rod  a,  and  communicate  the  charge  to  be 
measured  to  the  gilt  ball  m.  Replace  the  glass  rod  in  the 
jar.  The  two  gilt  balls  will  touch  each  other  momentarily 
and  half  of  the  charge  will  pass  from  m  to  n.  As  both  balls 
possess  similar  charges,  they  will  immediately  repel  each 
other;  the  ball  n  being  driven  around,  twists  up  the  wire  to 
a  certain  extent.  The  resistance  of  the  wire  to  torsion  will 
eventually  balance  the  force  of  repulsion,  and  the  ball  n  will 
come  to  rest,  the  balls  being  separated  by  a  certain  distance. 
In  any  wire,  the  force  of  torsion  is  proportional  to  the  amount 
of  twist,  or,  in  this  case,  to  the  angle  of  torsion ;  hence,  the 
force  exerted  between  the  two  balls  can  be  measured  by  the 
angle  described  by  the  ball  n. 

19O.  By  means  of  the  torsion  balance,  it  is  proved  that 
the  force  exerted  between  two  bodies,  statically  charged  witJi 
electricity,  varies  inversely  as  the  square  of  the  distance 
between  them. 

Thus,  suppose  two  electrified  bodies  \  inch  apart  repel 
each  other  with  a  certain  force ;  at  a  distance  of  1  inch,  the 
force  would  only  be  one-sixteenth  as  great.  This  law  is 
equally  true  for  the  force  of  attraction  between  two  bodies 
with  dissimilar  charges. 

A  unit  quantity  of  electricity  is  that  charge  which,  when 
placed  in  air  at  a  distance  of  1  centimeter  from  another  equal 
and  similar  charge,  will  be  repelled  with  the  force  of  1  dyne. 

NOTE. — (a)    The    centimeter,   or    the    unit    of    length,    represents 

of  the  distance  from  the  pole  to  the  equator  on  the  surface 
1,000,000,000 


§  4  PHYSICS.  107 

of  the  earth,  and  is  equal  to  .3937  inch,  or  1  inch  equals  2.54  centi- 
meters, nearly.  A  square  centimeter  is  the  area  contained  in  a  square 
each  of  whose  sides  is  1  centimeter  in  length ;  1  square  centimeter 
equals.  155 square  inch,  or  1  inch  equals  6.45  square  centimeters,  nearly. 
A  cubic  centimeter  is  the  volume  contained  in  a  cube  each  of  whose 
edges  is  1  centimeter  in  length;  1  cubic  centimeter  equals  .06102  cubic 
inch,  or  1  cubic  inch  equals  16.387  cubic  centimeters. 

(b)  The  gram,  or  the  unit  of  mass  or  quantity  of  matter,  represents 
the  quantity  of  matter  contained  in  a  cubic  centimeter  of  pure  water  at 
the  temperature  of  its  maximum  density,  which  is  4°  C.,  or  39.2°  F., 
and  is  equal  in  weight  to  15.432  grains. 

(c]  The  second,  or  the  unit  of  time,  represents  ^  of  an  ordinary 

day  of  24  hours. 

This  system  of  units  is  designated  as  the  absolute  or  C.  G.  S.  system, 
to  distinguish  it  from  other  systems  based  upon  other  fundamental  units. 

The  secondary  units,  derived  from  ^hese  fundamental  units,  are  the 
unit  of  velocity  and  the  dyne. 

The  unit  of  velocity,  or  the  rate  at  which  a  body  changes  its  relative 
position,  is  determined  by  dividing  the  distance  in  centimeters  through 
which  a  body  travels  by  the  time  in  seconds  required  to  travel  that 
distance.  The  unit  of  velocity  is,  therefore,  1  centimeter  per  second. 
To  compute  the  velocity  in  centimeters  per  second,  divide  the  num- 
ber of  centimeters  by  the  number  of  seconds. 

The  dyne,  or  the  unit  of  force,  is  that  force  which,  by  acting  upon  a 
mass  of  1  gram  for  1  second,  can  give  to  it  a  velocity  of  1  centimeter 
per  second. 

191.  In  either  case,  whether  of  attraction  or  repulsion, 
the  force  at  any  given  distance  is  equal  to  the  product  of  the 
two  quantities  of  electricity  on  the  bodies.     That  is,  if  a  cer- 
tain body  were  charged  with  4  unit  quantities  of  electricity, 
and  another  with  3  unit  quantities,  then  the  force  exerted 
between  them  would  be  12  times  as  great  as  if  each  had  con- 
tained a  charge  of  1  unit. 

192.  Only  that  part  of  a  dry  glass  rod  which  has  been 
rubbed   will   be   electrified;   the   other   parts   will   produce 
neither  attraction  nor  repulsion  when  brought  near  an  elec- 
troscope.    The  same  is  true  of  a  piece  of  sealing  wax  or 
resin.     These  bodies  do  not  readily  conduct  electricity,  that 
is,  they  oppose  or  resist  the  passage  of  electricity  through 
them.    Therefore,  electricity  can  reside  only  as  a  charge  upon 
that  part  of  their  surfaces  where  it  is  developed.     Experi- 
ments show  that  when  a  metal  receives  a  charge  at  any 
point,  the  electricity  immediately  passes  or  flows  through  its 
substance  to  all  parts.     Metals,   therefore,  are  said  to  be 


108  PHYSICS.  §  4 

good  conductors  of  electricity.  Bodies  have,  accordingly, 
been  divided  into  two  classes;  namely,  non-conductors  or 
insulators,  those  bodies  that  offer  an  infinitely  high  resistance 
to  the  passage  of  electricity ;  and  conductors,  or  those  that 
offer  a  comparatively  low  resistance  to  its  passage.  This 
distinction  is  not  absolute,  for  all  bodies  conduct  electricity 
to  some  extent,  while  there  is  no  known  substance,  that  does 
not  offer  some  resistance  to  its  flow. 

11)3.  Rlcctrical  resistance  maybe  defined  as  a  general 
property  of  matter,  varying  with  different  substances,  by 
virtue  of  which  matter  opposes  or  resists  the  passage  of  elec- 
tricity. 

1 94.  Conductivity  is  the  facility  with  which  a  body  trans- 
mits electricity,    and   is   the   opposite    of   resistance.       For 
instance,  copper  is  of  low  resistance  and  high  conductivity; 
wood  is  of  high  resistance  and  low  conductivity. 

195.  In  giving  the  following  list  and  dividing  the  differ- 
ent substances  into  two  classes,  it  should  be  understood  that 
it  is  done  only  as  a  guide  for  the  student.     Between,  these 
classes  are  many  substances  that  might  be  included  in  either, 
and  no  hard-and-fast  line  can  be  drawn.    The  list  is  arranged 
in  order  of  the  conductivity  of  the  different  substances,  begin- 
ning with  silver,  which  is  the  best  known  conductor. 

f  Paper  . 
Oils 
Porcelain 

Wood 

Silver  NON-CON- 

Copper  DUCTORS 


CON- 


Other  metals  OR 


DUCTORS   i  INSULA- 


Charcoal 
Ordinary  water 


TORS 


Silk 


Resins 

Gutta  percha 

Shellac 

Ebonite 

Paraffin 

Glass 

Dry  air 


§  4  PHYSICS.  109 

196.  Arr  electric  charge  will  be  induced  in  a  conductor 
when  that  conductor  is  brought  into  the  vicinity  of  an  elec- 
trified body.  This  effect  is  termed  electrostatic  induction, 
and  the  range  of  space  in  which  it  can  take  place  is  an  elec- 
trostatic field. 

If  the  conductor^  B,  Fig.  48,  is  supported  from  contact 
with  the  earth  by  insulators,  and  is  then  brought  into  the 
electrostatic  field  of  the 
conductor    C]    but    not 
touching    (7,    which    is 
electrified  with  ^.positive 
charge,     the     following 
facts  will  be  observed: 

1.  A  charge  will  be 
produced  on  A  B,  as  is 
shown  by  the  pith  balls 

...    J  FIG.  48. 

spreading  apart. 

2.  This  charge  will  be  negative  at  the  end  A  nearest  C, 
and  positive  at  the  end  B  farthest  from  C,  as  can  be  shown 
by  an  electroscope. 

3.  The  charges  at  A  and  B  are  equal  to  each  other;  for, 
if  the  conductor  A  B  is  removed  from  the  vicinity  of  the 
conductor  C  without  having  touched  C,  the  opposite  charges 
immediately  neutralize  each  other;  that   is,   no  electrifica- 
tion will  be  indicated  by  the  pith  balls. 

4.  Again,  as    C  is   brought   nearer   and   nearer   A,  the 
charges  on  opposite  signs  on  the  approaching  surfaces  attract 
each  other 'more  and  more  strongly  until  C  is  brought  very 
near  A,  when  a  spark  darts  across  the  intervening  space. 
The  two  charges,  rushing  together,  neutralize  each  other, 
leaving  the  induced  positive   charge,   which   was   formerly 
repelled  to  the  end   B  of  the  conductor,  as  a  permanent 
charge  over  all  the  surface  of  A  B. 

5.  Or,  if  the  conductor  A  B  is  touched  by  a  conductor 
connected  to  the  earth  when  it  is  under  the  influence  of  C, 
the  positive  charge  will  be  neutralized  by  the  earth,  and  the 
negative  charge  will  remain  when  A  B  is  removed  from  the 
field  of  C.     The  charge  that  passes  to  the  earth  from  A  B  is 


110  PHYSICS.  §  4 

called  a  free  charge,  while  that  charge  which  is  held  by  the 
inductive  influence  of  C  is  a  bound  cliarge.  Both  free  and 
bound  charges  can  be  negative  or  positive,  depending  upon 
the  sign  of  the  charge  on  C. 

When  two  conducting  bodies,  both  electrified  with  equal 
dissimilar  charges,  are  touched  together  momentarily,  the 
two  charges  will  neutralize  each  other,  no  trace  of  either 
remaining;  but  if  they  are  unequal,  the  smaller  charge  will 
neutralize  an  equal  amount  from  the  larger  and  leave  a 
charge  that  is  equal  to  the  difference  between  the  two 
original  charges,  the  sign  of  the  remaining  charge  being  the 
same  as  that  of  the  larger  one.  Before  the  bodies  can  be 
separated,  the  remaining  charge  will  divide  equally  between 
the  two  bodies.  For  example,  two  gilt  balls  A  and  B  are 
charged  respectively  with  -j-  20  and  —  4  units  of  electricity. 
When  the  balls  are  placed  in  contact,  the  —  4  charge  on  B 
will  neutralize  a  +4  charge  on  A  and  leave  a  -f  16  charge, 
which  immediately  divides  equally  between  the  two  balls; 
that  is,  a  charge  of  -j-  8  units  remains  on  each  ball  when  they 
are  separated. 

It  is  found  that  the  effect  of  this  electrostatic  induction  is 
greatly  increased  by  placing  some  other  substance,  such  as 
glass  or  paper,  between  the  two  bodies. 

197.  The  facility  with -which  a  body  allows  electrostatic 
induction  to  act  across  it  is  called  its   inductiiw  capacity. 
The  inductive  capacity  varies  with  different  substances,  but 
almost  all  non-conductors  are  better  than  air. 

198.  The   electrophorus,   Fig.    40,   is   an   instrument 
devised  for  the  purpose  of  obtaining  an  almost  unlimited 
number   of   static   charges    of   electricity   from   one    single 
charge,   and   is   based    upon    the   principle    of   electrostatic 
induction. 

It  consists  of  two  main  parts:  a  thin  cake  of  resinous 
material  cast  in  a  round  metal  dish  or  pan  B,  about  1  foot  in 
diameter;  and  a  round  disk  A  of  slightly  smaller  diameter, 
made  of  metal  and  provided  with  a  glass  handle.  In  using 


PHYSICS. 


Ill 


the  electrophorus,  the  resinous  cake  must  first  be  beaten  or 
rubbed  with  a  warm  piece  of  woolen  cloth  or  fur.  The 
disk,  or  cover,  is  then  placed  on  the  cake,  touched  momen- 
tarily with  the  finger  to  liberate  the  free  charge,  then 
removed  by  taking  up  by  the  handle.  It  is  now  found  to  be 
powerfully  electrified  with  a  positive  charge;  so  much  so, 
indeed,  as  to  yield  a  considerable  spark  when  the  hand  is 


FIG.  49. 

brought  near  it.  The  cover  may  be  replaced,  touched,  and 
again  removed,  and  will  thus  yield  any  number  of  sparks; 
the  original  charge  on  the  resinous  plate  meanwhile  remain- 
ing practically  as  strong  as  ever. 

199.  A  static  charge  of  electricity  is  not  usually  distrib- 
uted uniformly  over  the  surface  of  conducting  bodies. 
Experiments  show  that  there  is  more  electricity  on  the 
edges  and  corners  than  upon  their  flatter  parts. 

The  term  electric  density  is  used  to  signify  the  quantity  of 
electricity  residing  in  a  small  area  at  any  part  of  a  body, 
the  distribution  being  supposed  to  be  imiform  over  that 
small  part  of  the  surface. 

The  electric  density  is  the  quotient  arising  from  dividing  the 
total  charge  of  electricity  in  units  of  quantity  residing  upon 
the  surface  of  a  body,  by  the  area  of  the  surface  in  square 
inches.  For  example,  a  charge  of  240  units  of  electricity  is 


112 


PHYSICS. 


imparted  to  a  sphere,  the  surface  area  of  which  is  40  square 
inches;  then,  the  electric  density  over  the  surface  of  the 
sphere  is  -2T\°-  =  6  units  of  electricity  per  square  inch. 


200.  Electrostatic   machines  have  been  devised  for 
the  purpose  of  obtaining  larger  static  charges  than  can  be 
developed  by  rubbing  the  glass  rod  or  by  the  electrophorus. 
They  consist,  mainly,  of  two  parts,  one  for  producing  and 
the  other  for  collecting  the  charges. 

There  are  three  important  kinds  of  electrostatic  machines : 

the  cylinder,  the  plate ',  and  the  induction  machines. 

0 

20 1.  The  cylinder  machine,  as  usually  constructed,  con- 
sists of  three  principal  parts:  a  cylinder  of  glass,  revolving 
upon  a  horizontal  axis ;  a  rubber,  or  cushion  of  horsehair, 
to  which  is  attached  a  long  silk    flap;    and  an   insulating 
metallic  cylinder  called  a  prime  conductor.     In  Fig.  50,  the 


FIG.  50. 

cushion  of  horsehair  a,  covered  with  a  coating  of  amalgam 
of  zinc,  presses  against  the  glass  cylinder  b  from  behind, 
allowing  the  silk  flap  s  to  rest  upon  the  upper  half  of  the 
glass.  The  prime  conductor  C  is  provided  at  one  end  with 
a  row  of  fine  metallic  spikes  and  is  placed  in  front  of  the 
machine  with  the  row  of  spikes  projecting  towards  the  glass 
cylinder.  When  the  glass  cylinder  is  rotated,  a  positive 
charge  is  produced  upon  the  glass  and  a  negative  charge  upon 
the  rubber.  The  positive  charge  is  carried  around  upon  the 
glass  cylinder,  and,  just  before  reaching  a  position  opposite  the 
row  of  spikes,  it  acts  inductively  upon  the  prime  conductor, 


§4  PHYSICS.  113 

attracting  a  negative  charge  to  the  near  end  and  repell- 
ing a  positive  charge  to  the  far  end.  When  the  positive 
charge  arrives  in  front  of  the  row  of  spikes,  it  will  be  neu- 
tralized by  the  attracting  negative  charge  from  the  con- 
ductor, leaving  the  glass  in  a  neutral  condition,  ready  to  be 
excited  again.  A  positive  charge  now  remains  upon  the 
prime  conductor  and  can  be  utilized  for  other  experiments. 

202.  The  plate  machine  is  similar  in  all  respects  to  the 
cylinder  machine  with  the  exception  that  a  glass  or  ebonite 
plate  is  used  instead  of  the  glass  cylinder,  and  there,  are 
usually  two  sets  of  rubbers  or  cushions,   instead   of   c>ne. 
Each  set  of  cushions  is  double;  that  is,  it  is  made  in  two 
parts,  with  the  plate  revolving  between  them.     One  set  of 
cushions  is  placed  at  the  top  of  the  machine,  and  the  other 
at  the  bottom,  with  silk  flaps  extending  from  each  over  a 
quadrant  of  the  plate.     The  charge  is  collected  on  two  prime 
conductors  connected  by  a  metal  rod,  and  each  is  provided 
with  a  row  of  fine  spikes  at  one  end.     They  are  placed  in 
such  a   position   that  the  two  rows  of   fine  spikes   project 
towards  the  glass  plate  at  opposite  sides  of  its  horizontal 
diameter.     The  electrostatic  action  of  the  machine  is  in  all 
respects  the  same  as  that  of  the  cylinder  machine. 

203.  The  induction  machine  differs  widely  in  its  action 
from  the  two  machines  previously  described.     It  requires  an 
initial  charge  from  some  exterior  source  to  start  its  action. 
The  initial  charge  acts  inductively  across  a  revolving  glass 
plate,  and  produces  other  charges;   these   charges  in  turn 
are  conveyed  by  the  moving  parts  to  some   other   point, 
where  they  increase  the  initial  charge,  or  furnish  a  supply 
of  electricity  to  a  prime  conductor. 

The  two  principal  machines  of  this  class  are  the  Holtz  and 
the  Wimshurst. 

204.  It  has  been  shown  that  opposite  charges  attract 
and  hold  one  another;  that  electricity  cannot  flow  through 
glass,  and  yet  can  act  across  it  by  induction.     If  a  piece  of 
tin  foil  is  pasted  upon  the  middle  of  each  face  of  a  thin  plate 


114  PHYSICS.  §  4 

of  glass,  and  one  of  the  pieces  is  electrified  with  a  positive 
charge  and  the  other  with  a  negative  charge,  the  two 
charges  will  attract  each  other  ;  or,  in  other  words,  they  are 
held,  or  bound,  by  each  other.  It  will  be  found  that  these 
two  pieces  of  tin  foil  may  be  charged  a  great  deal  stronger 
in  this  manner  than  either  of  them  could  possibly  be  if  they 
were  stuck  to  the  glass  alone  and  then  electrified.  This 
property  of  retaining  and  accumulating  a  large  quantity  of 
static  charges,  which  two  conductors  possess  when  placed 
side  by  side  and  separated  from  each  other  by  a  non-con- 
ductor, is  termed  capacity. 

2O5.  A  condenser  is  an  apparatus  for  condensing  or 
accumulating  a  large  quantity  of  static  charges  of  electricity 
on  a  comparatively  small  surface.  It  consists  of  two  con- 
ductors separated  by  a  thin  layer  of  some  non-conducting 
material;  one  of  the  plates  is  entirely  insulated  from  the 
earth,  and  the  other  is  connected  to  it  by  a  conductor. 

The  capacity  of  a  condenser  depends  upon  the  size  and 
form  of  the  condensing  plates,  the  thinness  of  the  insula- 
ting material  between  them,  and  the  inductive  capacity  of 
the  insulating  material. 


A  convenient  form  of  condenser  is  called  the  Ley- 
den  jar,  Fig.  51.     It  consists  of  a  glass  jar  J  coated  up  to 


FIG.  51. 


a  certain  height  on  the  inside  and  outside  with  tin  foil.  A 
brass  knob  a  is  fixed  on  the  end  of  a  stout  brass  wire,  which 
passes  downwards  through  a  lid  or  stopper  of  dry,  well 


§4  PHYSICS.  115 

varnished   wood,  and  connected  by  a  loose   piece   of  brass 
chain  with  the  inner  coating  of  the  jar. 

To  charge  the  jar,  the  knob  is  held  to  the  prime  conductor 
C  of  an  electrical  machine,  the  jar  being  either  held  in  the 
hand  by  the  outer  tin  foil  coating  or  connected  to  the  earth 
by  a  wire  or  chain.  When  a  positive  charge  is  thus  imparted 
to  the  inner  coating,  it  acts  inductively  on  the  outer  coating, 
attracting  a  negative  charge  in  the  face  of  the  outer  coating 
nearest  the  glass,  and  repelling  a  positive  charge  to  the  out- 
side of  the  outer  coating.  This  outer  charge  then  passes 
through  the  hand  or  any  conductor  to  the  earth. 

2O  7.  An  electrostatic  battery  consists  of  a  number 
of  Leyden  jars  whose  inside  coatings  are  all  connected 
together  and  whose  outside  coatings  are  all  connected  to  the 
earth. 


ELECTRODYNAMICS. 

208.  In  dealing  with  electric  currents  the  word  potential 
will  be  substituted  for  the  general  and  vague  phrase  electrical 
condition. 

The  term  potential,  as  used  in  electrical  science,  is  analo- 
gous to  pressure  in  gases,  head  in  liquids,  and  temperature 
in  heat. 

When  an  electrified  body,  positively  charged,  is  connected 
to  the  earth  by  a  conductor,  electricity  is  said  to  flow  from 
the  body  to  the  earth;  and,  conversely,  when  an  electrified 
body  negatively  charged  is  connected  to  the  earth,  electricity 
is  said  to  flow  from  the  earth  to  that  body.  That  which 
determines  the  direction  of  floiv  is  the  relative  electrical 
potential  or  pressure  of  the  two  charges  with  respect  to  the 
earth. 

209.  It  is  impossible   to  say  with  certainty  in  which 
direction    electricity    really   flows,    or,   in    other  words,    to 
declare  which  of  two  points  has  the  higher  and  which  the 


116  PHYSICS.  §  4 

lower  electrical  potential,  or  pressure.  All  that  can  be  said 
with  certainty  is,  that  when  there  is  a  difference  of  electrical 
potential,  or  pressure,  an  electric  current  tends  to  flow/r<?;// 
the  point  of  higher  to  that  of  lower  potential,  or  pressure. 

For  convenience,  it  has  been  arbitrarily  assumed  and 
universally  adopted,  that  that  electrical  condition  called 
positive  is  at  a  higher  potential,  or  pressure,  than  that  called 
negative;  and  that  an  electric  current  flows  from  a  positively 
to  a  negatively  electrified  body. 

2  1C.  The  zero,  or  normal,  level  of  water  is  taken  as 
that  of  the  surface  of  the  sea,  and  the  normal  pressure  of  air 
as  that  of  the  atmosphere  at  the  sea  level  ;  similarly,  there  is 
a  zero  pressure,  ox  potential,  of  electricity  in  the  earth  itself. 
It  may  be  regarded  as  a  reservoir  of  electricity  of  infinite 
quantity,  and  its  pressure,  or  potential,  taken  as  zero.  In 
Art.  2O8,  the  condition  called  positive  is  assumed  to  be  at 
a  higher  potential  than  the  earth,  and  that  called  negative  is 
assumed  to  be  at  a  lower  potential  than  the  earth. 

It  must  be  understood  that  electricity  is  a  condition  of 
matter  and  not  matter  itself,  for  it  possesses  neither  weight 
nor  extension.  Consequently,  the  statement  that  electricity 
is  flowing  through  a  conductor  must  not  be  taken  too  liter- 
ally; it  must  not  be  supposed  that  any  material  substance, 
such  as  a  liquid,  is  actually  passing  through  the  conductor 
in  the  same  sense  as  water  flows  through  a  pipe.  The  state- 
ment that  electricity  is  flowing  through  a  conductor  is  only 
another  way  of  expressing  the  fact  that  the  conductor  and 
the  space  surrounding  it  are  in  different  conditions  than  usual, 
and  that  they  possess  unusual  properties.  The  action  of 
electricity,  however,  is  quite  similar  in  many  respects  to  the 
flow  of  liquids,  and  the  study  of  electric  currents  is  much 
simplified  by  the  analogy. 


In  order  to  produce  what  is  called  an  electric  cur- 
rent, it  is  first  necessary  to  cause  a  difference  of  electrical 
potential,  or  pressure,  between  two  bodies  or  between  two  parts 
of  the  same  body. 


§  4  PHYSICS.  117 

In  Art.  186,  it  was  stated  that  when  two  dissimilar  sub- 
stances are  simply  placed  in  contact,  one  always  assumes 
the  positive  and  the  other  the  negative  condition ;  in  other 
words,  a  difference  of  electrical  potentially  developed  between 
the  two  bodies. 

Placing  a  piece  of  copper  and  zinc  in  contact  will  develop 
a  difference  of  electrical  potential,  which  can  easily  be 
detected.  The  same  results  will  follow  if  the  plates  are 
slightly  separated  from  each  other  and  placed  in  a  vessel 
containing  saline  or  acidulated  water,  leaving  a  small  portion 
of  one  end  of  each  plate  exposed.  The  exposed  ends  of  the 
zinc  and  copper  are  now  electrified  to  different  degrees;  or, 
in  other  words,  there  is  a  difference  of  electrical  potential 
between  the  plates,  one  plate  being  at  a  higher  potential 
than  the  other. 

When  the  exposed  ends  are  connected  together  by  any 
conducting  material,  the  potential  between  the  plates  tends 
to  equalize,  and  a  momentary  rush  or  discharge  of  electricity 
passes  between  the  exposed  ends  through  the  conducting 
material  and  between  the  submerged  ends  through  the  liquid. 
During  its  passage  through  the  liquid,  the  electricity  causes 
certain  chemical  changes  to  take  place ;  these  chemical  reac- 
tions cause  in  their  turn  a  fresh  difference  of  potential 
between  the  plates,  which  is  followed  immediately  by  another 
equalizing  discharge,  and  that  by  a  further  difference,  and 
so  on.  These  changes  follow  one  another  with  great  rapid- 
ity; so  rapid,  in  fact,  that  it  is  impossible  to  distinguish 
them,  and  they  appear  absolutely  continuous.  The  equalizing 
flow  that  is  constantly  taking  place  from  one  plate  to  the 
other  is  known  as  a  continuous  current  of  electricity.  Con- 
sequently, an  electric  current  becomes  continuous  when  the 
difference  of  potential  is  constantly  maintained. 

By  the  use  of  a  very  delicate  electroscope,  the  exposed  end 
of  the  copper  will  be  found  to  be  electrified  with  a  positive 
charge,  and  the  submerged  end  with  a  negative  charge ;  in 
the  case  of  the  zinc,  the  .opposite  conditions  exist,  namely, 
the  exposed  end  is  electrified  with  a  negative  charge,  and 
the  submerged  end  with  a  positive  charge.  The  current, 


118  PHYSICS.  §  4 

therefore,  will  flow  from  the  exposed  end  of  the  copper 
through  the  conductor  to  the  exposed  end  of  the  zinc,  and 
from  the  submerged  end  of  the  zinc  through  the  liquid  to 
the  submerged  end  of  the  copper. 

*-  1  '*.     A  simple  voltaic  or  galvanic  cell  is  shown  in 

Fig.  52.  It  is  an  apparatus  for 
developing  a  continuous  current 
of  electricity,  and  consists,  essen- 
tially, of  a  vessel  A  containing 
saline  or  acidulated  water,  in 
which  are  submerged  two  plates 
of  dissimilar  metals,  C  and  Z,  or 
one  metal  and  a  metalloid. 

Electrolyte  is  the  name  given 
to  the  liquid  that,  as  it  trans- 
mits the  current,  is  decomposed 
by  it. 

The    two    dissimilar     metals, 

when  spoken  of  separately,  are  called  voltaic  elements;  when 
taken  collectively,  they  are  known  as  a  voltaic  couple. 

A  voltaic  battery  is  a  number  of  simple  voltaic  cells  properly 
joined  together. 

Electrodes,  or  poles,  of  a  cell  or  battery  are  metallic  ter- 
minals attached  to  the  plates,  and  are  used  to  connect  the 
cell  or  battery  to  any  exterior  conductor  or  to  another  cell 
or  battery. 

It  should  be  remembered  that  the  polarity  of  that  end  of 
the  plate  or  voltaic  element,  which  is  acted  upon  by  the 
electrolyte,  is  always  of  opposite  sign  to  its  electrode.  For 
instance,  in  the  case  of  the  zinc  and  copper,  the  electrode 
fastened  to  the  zinc  would  be  spoken  of  as  the  negative 
electrode  of  the  cell;  while  the  zinc  itself  would  be  the 
positive  element  of  the  cell,  its  submerged  end  being 
positive. 

213.  Chemical  Actions  That  Take  Place  in  a  Simple 
Cell. — When  a  piece  of  ordinary  zinc  is  placed  alone  in 


§  4  PHYSICS.  119 

sulphuric  acid  diluted  with  water,  the  zinc  is  attacked  by  the 
acid  and  a  part  of  it  is  dissolved  into  a  salt  of  that  metal 
called  sulpJiate  of  zinc.  At  the  same  time  the  liquid  is  decom- 
posed and  hydrogen  gas  is  liberated  from  it,  coming  up  from 
around  the  zinc  in  small  bubbles,  and  the  whole  mass  of  the 
liquid  becomes  heated.  If  the  zinc  is  absolutely  pure,  the 
chemical  actions  take  place  more  slowly;  the  bubbles  of 
hydrogen  do  not  immediately  rise  to  the  surface,  but  form 
around  the  zinc,  protecting  it  from  further  action  of  the 
acid.  By  placing  another  metal  in  the  water,  say  a  piece  of 
copper,  and  connecting  its  exposed  end  with  that  of  the  zinc 
by  a  conductor,  the  chemical  actions  become  exceedingly 
vigorous  again.  Large  quantities  of  hydrogen  gas  are 
again  liberated,  but  instead  of  the  bubbles  appearing 
around  the  zinc,  they  form  around  the  'copper  and  come 
to  the  surface  at  that  place ;  the  energy,  which  in  the 
former  case  was  expended  in  heating  the  liquid,  now  ap- 
pears in  the  form  of  electric  energy.  Whenever  the  con- 
nection between  the  exposed  ends  is  broken,  all  chemical 
actions  cease  and  remain  inactive  until  the  two  metals  are 
again  connected. 

In  any  voltaic  cell,  the  element  that  is  acted  upon  by 
the  electrolyte  will  always  be  the  positive  element,  and  its 
electrode  the  negative  electrode  of  the  cell. 

214.  The  following  list  of  voltaic  elements  compose  the 
electromotive  series: 


1.   Zinc     . 

5.   Iron 

10.   Silver 

2.   Cadmium 

6.   Nickel 

11.   Gold 

3.   Tin 

7.   Bismuth 

12.   Platinum 

4.   Lead 

8.   Antimony 
9.   Copper 

13.   Graphite 

Any  two  of -these  metals  form  a  voltaic  couple,  and  produce 
a  difference  of  potential  when  submerged  in  saline  or  acidu- 
lated water,  the  one  standing  first  on  the  list  being  \\\Q  positive 
element,  or  plate,  and  the  other,  the  negative.  For  example, 
if  nickel  and  graphite  are  used,  the  nickel  will  be  acted  upon 


120  PHYSICS.  §  4 

by  the  liquid  and  will  form  tine positive  element;  but  if  nickel 
and  zinc  are  used,  the  zinc  will  be  acted  upon  by  the  liquid, 
and  will  form  \he  positive  element. 

The  difference  of  potential  will  be  greater  in  proportion 
to  distance  between  the  positions  of  the  two  substances  in 
the  list.  For  example,  the  difference  of  potential  developed 
between  zinc  and  graphite  is  much  greater  than  that  devel- 
oped between  zinc  and  nickel ;  in  fact,  the  difference  of 
potential  developed  between  zinc  and  graphite  is  equal  to 
the  difference  of  potential  developed  between  zinc  and 
nickel,  plus  that  developed  between  nickel  and  graphite. 

215.  Electricity  flowing  as  a  current  differs  from  static 
charges  in  three  important  degrees :   its  potential  is  much 
lower,  its  actual  quantity  is  larger,  and  it  is  continuous. 

A  strong  voltaic  battery  of  several  cells  produces  only  a 
slight  effect  upon  a  gold-leaf  electroscope,  and,  apparently, 
none  of  its  parts  possess  the  property  of  attracting  light 
substances.  The  potential  of  a  current  of  electricity  is  com- 
paratively so  small  that  a  voltaic  battery  composed  of  a  large 
number  of  cells  is  not  sufficient  to  produce  a  spark  of  more 
than  y^  or  -5-^  inch  in  air,  whereas  a  small  electrostatic 
machine  will  produce  sparks  several,  inches  in  length.  If, 
however,  the  actual  quantity  of  electricity  is  measured  by 
its  effects  in  decomposing  water,  then  the  quantity  produced 
by  a  simple  voltaic  cell  as  small  as  a  thimble  would  give 
greater  results  than  that  from  an  electrostatic  machine  with 
plates  2  or  3  feet  in  diameter. 

An  electric  current  cannot  be  developed  upon  the  surfaces 
of  non-conducting  substances  by  current  electricity,  as  in  the 
case  of  static  charges,  and  it  will  never  flow  unless  the  con- 
ducting path  is  made  entirely  of  conducting  material. 

216.  A  number  of  contacts  of  dissimilar  metals  can  be 
so  arranged  as  to  add  their  electrical  effects  together;  the 
difference  of  potential  then  developed  will   be   greater  in 
proportion  to  the  number  of  contacts.      Such  an  arrange- 
ment is  called  a  voltaic  pile.      (See  Fig.   53.)    It  is  made  by 


PHYSICS. 


121 


placing  a  pair  of  disks  of  zinc  and  copper  in  contact  with  each 
other,  and  then  laying  a  piece  of  flannel  or  blotting  paper, 
moistened  with  brine,  upon  the  copper  disk.  The  pair  of 
disks  now  form  a  voltaic  couple.  Several  voltaic  couples 
are -placed  together  and  each  pair  sepa- 
rated by  a  moistened  piece  of  flannel  or 
blotting  paper.  One  end  of  such  a  pile 
would  then  be  terminated  by  a  disk  of 
copper  and  the  other  by  a  disk  of  zinc. 
The  copper  forms  the  positive  electrode 
and  the  zinc  the  negative  electrode. 
When  these  two  electrodes  are  joined 
by  a  conductor,  a  current  will  flow  from 
the  positive  to  the  negative  through  the 
conductor,  and  from  the  negative  to  the 
positive  through  the  contacts. 

217.  The  difference  of  potential 
developed  by  the  mere  contact  of  two 
dissimilar  metals  varies,  not  only  with 
the  metals  employed  and  the  physical 
condition  of  each,  but  also  with  their 
temperature. 

The  greater  difference  of  potential 
developed  by  heat  can  be  shown  by 
soldering  one  end  of  a  bar  of  copper  to  one  end  of  a  bar  of 
zinc,  and  applying  heat  to  the  juncture  so  as  to  raise  its 
temperature  above  that  of  the  other  parts  of  the  bars. 
By  joining  the  free  ends  together  with  a  conductor,  a  cur- 
rent of  electricity  will  flow  from  the  zinc  through  the  con- 
tact to  the  copper;  then  from  the  free  end  of  the  copper 
to  the  free  end  of  the  zinc  through  the  conductor.  If  the 
junction  be  cooled  below  the  other  parts  of  the  bars,  a 
current  is  produced  in  the  opposite  direction ;  that  is, 
from  the  copper  through  the  contact  to  the  zinc,  etc.  Even 
the  same  metal  in  different  physical  conditions  will  develop 
a  difference  of  potential  if  heated  in  a  certain  place.  For 
instance,  take  a  copper  wire,  part  of  which  is  straight  and 


FIG.  53. 


122  PHYSICS.  §  4 

the  remainder  bent  into  a  spiral,  and  heat  the  place  where 
the  spiral  begins.  Under  these  conditions,  a  difference 
of  electrical  potential  will  be  developed  between  the  two 
free  ends. 

In  general,  the  difference  of  potential  is  larger  in  propor- 
tion as  the  difference  of  temperature  increases.  With  extreme 
temperature,  however,  this  condition  changes,  and  at  a  cer- 
tain temperature  of  the  junction  no  difference  of  potential 
whatever  is  noticed.  This  temperature  is  called  the  neutral 
temperature.  When'  the  junction  is  heated  beyond  the  neu- 
tral temperature,  inversion  takes  place;  that  is,  the  direction 
of  the  current  changes. 

Electric  currents  produced  by  a  change  of  temperature  are 
called  thermoelectric  currents. 

On  account  of  the  small  difference  of  potential  of  thermo- 
electric currents,  they  are  of  no  practical  value ;  in  fact,  they 
become  a  source  of  great  annoyance  and  error  in  accurate 
measurements  with  delicate  instruments. 


ELECTKOMAGKETISM. 


218.  If  a  conductor  conveying  a  current  of  electricity  is 
brought  near  a  freely  suspended  magnetic  needle,  the  needle 
tends  to  place  itself  at  right  angles  to  the  conductor,  as 
shown  in  Fig.  54 ;  or,  in  general,  an  electric  current  and  a 


FIG.  54. 


magnet  exert  mutual   force  upon   each   other.     From   the 
definition   given   in  Art.  174,  the   space   surrounding  the 


PHYSICS. 


123 


conductor  is  a  magnetic  field.  If  the  conductor  is  threaded 
up  through  a  piece  of  cardboard, 
and  iron  filings  are  sprinkled  on 
the  cardboard,  they  will  arrange 
themselves  in  concentric  circles 
around  the  conductor,  as  repre- 
sented in  Fig.  55.  This  effect  will 
be  observed  throughout  the  entire 
length  of  the  conductor,  and  is 
caused  entirely  by  the  current.  In 
fact,  every  conductor  conveying  a 
current  of  electricity  can  be  imag- 
ined as  completely  surrounded  by 
a  sort  of  magnetic  wliirl,  the  magnetic  density  decreasing 
as  the  distance  from  the  current  increases  (see  Fig.  56). 

If  the  current  in  a  horizontal  conductor  is  flowing 


FIG.  55. 


FIG.  56. 


towards  the  north  and  a  compass  is  placed  under  the  con- 
ductor, Fig.  57,  the  north  pole  of  the  needle  will  be  deflected 
towards  the  west ;  by  placing  the  compass  over  the  wire, 


FIG.  57. 


FIG.  58. 


FIG.  59. 


Fig.  58,  the  north  pole  of  the  needle  will  be  deflected  to- 
wards the  east.     By  reversing  the  direction  of  the  current  in 


124 


PHYSICS. 


the  conductor,  the  needle  will  point  in  the  opposite  direction 
in  each  case,  respectively. 

If  the  conductor  is  placed  over  the  needle  and  then  bent 
back  under  it,  forming  a  loop,  as  shown  in  Fig".  59,  the 
tendency  of  the  current  in  both  top  and  bottom  portions  of 
the  wire  is  to  deflect  the  north  pole  of  the  needle  in  the 
same  direction. 

From  these  experi- 
ments, knowing  the  direc- 
tion of  current  in  the  con- 
ductor, the  following  rule 
is  deduced  for  the  direc- 
tion of  the  lines  of  force 
around  the  conductor: 

Rule. — If  tJie  current 
J  is  floiving  in  tJie  conductor 
away  front  the  observer, 
tlicn  the  direction  of  the 
lines  of  force  unit  be 
around  tJie  conductor  in 
tJie  direction  of  the  hands- 
of  a  watcli. 

The  direction  of  the 
lines  of  force  around  a 
conductor  is  indicated  in  Fig.  60,  where  the  current  is  assumed 
to  be  flowing  downwards;  that  is,  piercing  the  paper. 


FIG.  60. 


FIG.  61 


FIG.  62. 


2  2O.     Two   parallel  conductors,   both  transmitting  cur- 
rents of  electricity,  are  either  mutually  attractive  or  repellent, 


§  4  PHYSICS.  125 

depending  upon  the  relative  direction  of  their  currents.  If 
the  currents  are  flowing1  in  the  same  direction  in  both  con- 
ductors, as  represented  in  Fig.  61,  the  lines  of  force  will 
tend  to  surround  both  conductors  and  contract,  thus  attract- 
ing the  conductors.  If,  however,  the  currents  are  flowing 
in  opposite  directions,  as  in  Fig.  62,  the  lines  of  force  lying 
between  the  conductors  will  have  the  same  direction  and 
therefore  repel  the  conductors. 

22~L.  If  the  conductor  carrying  the  current  is  bent  into 
the  form  of  a  loop,  as  in  Fig.  63,  then  all  the  lines  of  force 
around  the  conductor  will  thread  through  the  loop  in  the 
same  direction.  By  bending  the  conductor  into  a  long  helix 
of  several  loops,  the  lines  of  force  around  each  loop  will  coin- 
cide with  those  around  the  adjacent  loops,  forming  several 
long  lines  of  force,  which  thread  through  the  entire  helix, 
entering  at  one  end  and  passing  out  at  the  other.  The  same 
conditions  now  exist  in  the  helix  as  exist  in  a  bar  magnet ;  i.  e. , 
the  lines  of  force  pass  out  from  one  end  and  enter  the  other. 


+       - 

/O 
// 


FIG.  63. 


In  fact,  the  helix  possesses  a  north  and  a  south  pole,  a  neu- 
tral line,  and  all  the  properties  of  attraction  and  repulsion  of 
a  magnet.  If  it  is  suspended  in  a  horizontal  position  and 
free  to  turn,  it  will  come  to  rest  pointing  in  a  north-and- 
south  direction.  A  helix  made  in  this  manner,  around  which 
a  current  of  electricity  is  circulating,  is  called  a  solenoid. 
The  polarity  of  a  solenoid — that  is,  the  direction  of  the 


126  PHYSICS.  §  4 

lines  of  force'that  thread  through  it — depends  upon  the  direc- 
tion in  which  the  conductor  is  coiled  and  the  direction  of  the 
current  in  the  conductor. 

To  determine  the  polarity  of  a  solenoid,  knowing  the  direc- 
tion of  the  current  : 

Rule. — In  looking  at  the  end  of  the  helix,  if  it  is  so  wound 
that  the  current  circulates  around  the  helix  in  the  direction 
of  the  liands  of  a  ivatch,  tliat  end  will  be  a  south  pole  ;  if  in 
tlie  other  direction  it  will  be  a  north  pole. 

Fig.   64  represents  a  conductor  coiled  in  a  right-handed 

helix.  If  the  current  starts 
to  flow  from  the  end  where 
the  observer  stands,  that  end 
will  be  a  south  pole  and  the 
observer  will  be  looking 
through  the  helix  in  the 
direction  of  the  lines  of  force. 

The  polarity  of  a  solenoid  can  be  changed  by  reversing  the 
direction  of  the  current  in  the  conductor. 

222.  In  Art.  176  it  was  stated  that  when  a  magnetic 
substance  is  brought  into  a  magnetic  field,  the  lines  of  force 
in  that  field  crowd  together  and  all  try  to  pass  through  that 
substance;  in  fact,  they  will  alter  their  circular  shape  and 
extend  a  considerable  distance  from  their  original  position 
in  order  to  pass  through  it.  A  magnetic  substance,  there- 
fore, offers  a  better  path  for  the  lines  of  force  than  air  or 
other  non-magnetic  substances. 

The  facility  afforded  by  any  substance  to  the  passage 
through  it  of  lines  of  force  is  called  magnetic  permeability,  or 
simply,  permeability. 

The  permeability  of  all  non-magnetic  substances,  such  as 
air,  copper,  wood,  etc.,  is  taken  as  1,  or  unity.  The  permea- 
bility of  soft  iron  may  be  as  high  as  2, 000  times  that  of  air. 
If,  therefore,  a  piece  of  soft  iron  is  inserted  into  the  mag- 
netic circuit  of  a  solenoid,  the  number  of  lines  of  force  will 
be  greatly  increased  and  the  iron  will  become  highly  mag- 
netized. 


PHYSICS. 


223.  The  Electromagnet. — A  magnet  produced  by 
inserting  a  magnetic  substance  into  the  magnetic  circuit  of  a 
solenoid  is  an  electromagnet,  and  a  magnetic  substance  around 
which  the  current  circulates 


FIG.  65. 


is  called  the  core  (see  Fig. 
65).  The  solenoid  is  gener- 
ally termed  the  magnetising 
coil. 

In  the  ordinary  form  of 
electromagnet,  the  magneti- 
zing coil  consists  of  a  large 
number  of  turns  of  insulated  wire,  that  is,  wire  covered  with 
a  layer  or  coating  of  some  non-conducting  or  insulating 
material,  usually  silk  or  cotton ;  otherwise,  the  current  would 
take  a  shorter  and  easier  circuit  from  one  coil  to  the  adjacent 
one,  or  from  the  first  to  the  last  coil  through  the  iron  core 
without  circulating  around  the  magnet. 


224. 

magnet. 


The  simplest  form  of  an  electromagnet  is  the  bar 
As  usually  constructed,  it  consists  of  a  straight  bar 
of  iron  or  steel  fitted  into  a  spool  or 
bobbin  made  of  hard,  vulcanized 
rubber  or  some  other  inflexible 
insulating  material.  The  magneti- 
zing coil  of  fine,  insulated  copper 
wire  is  wound  in  layers  in  the  bob- 
bin as  shown  by  the  cross-section 
in  Fig.  66. 

The  rule    for    determining  the 
polarity  of  a  solenoid,   Art.  £31, 
FlG-  66-  is  the  same  for  an  electromagnet. 

It  makes  no  difference  whether  the  wire  is  wound  in  one 
layer  or  in  any  number  of  layers,  or  whether  it  is  wound 
towards  one  end  and  then  wound  back  again  over  the 
previous  layer  towards  the  other  end;  so  long  as  the 
current  circulates  continually  in  the  same  direction  around 
the  core,  the  polarity  of  the  magnet  will  remain  un- 
changed. 


128 


PHYSICS. 


225.  The  most  convenient  form  of  an  electromagnet  for 
a  great  variety  of  uses  is  the  horseshoe  or  \}-shaped  electro- 
magnet. It  consists  of  a  bar  of  iron  bent  into  the  shape  of 
a  horseshoe  with  straight  ends,  and  is  provided  with  two 
magnetizing  coils,  one  on  each  end  of  the  magnet.  The  two 

ends  that  are  surrounded 
by  the  coils  are  the  cores  of 
the  magnet,  and  the  arc- 
shaped  piece  of  iron  join- 
ing them  together  is 
known  as  the  yoke  of  the 
magnet.  The  ordinary 
U-shaped  electromagnet 
is  made  in  three  parts; 
namely,  two  iron  cores 
wound  with  the  magnetizing  coils,  and  a  straight  bar  of  iron 
joining  thp  two  cores  together  for  a  yoke,  as  shown  in  Fig.  67. 
In  looking  at  the  free  ends  of  the  two  cores,  Fig.  68,  the 
current  should  circulate  around  one  core  in  an  opposite 

direction  to  that  around  the 
other.  If  the  current  circu- 
lates around  both  cores  in  the 


FIG.  67. 


FIG.  G8. 


FIG.  69. 


same  direction,  the  lines  of  force  produced  in  the  two  cores 
oppose  each  other,  respectively,  forming  two  like  poles  at 
their  free  ends  and  a  consequent  pole  in  the  yoke.  The  total 
number  of  lines  of  force  prodiiced  by  both  coils  will  be 


§  4  PHYSICS.  129 

greatly  diminished  and  the  mag-net  will  exhibit  only  a  small 
amount  of  magnetic  attraction. 

226.  Another  common  form  of  electromagnet  is  known 
as  the  iron-clad  electromagnet.  In  its  simplest  form,  Fig.  69, 
it  contains  only  one  magnetizing  coil  and  one  core.  The 
core  is  fastened  to  a  disk-shaped  iron  yoke,  and  the  mag- 
netic circuit  is  completed  through  an  iron  shell  that  rises  up 
from  the  yoke  and  completely  surrounds  and  protects  the 
coil. 


THEORETICAL  CHEMISTRY. 


INTRODUCTORY. 


MATTER  AND  FORCE. 

1.  Matter. — Although    the    objects  we    see   around    us 
differ  one  from  another  to  a  great  extent  in  appearance, 
shape,  and  character,  they  all  possess  one  property  in  com- 
mon, and  that  is  weight.     All  objects  are  attracted  by  the 
earth,  and  the  reason  why  we  find  a  body  to  be  heavy  is  that 
the  attraction  of  the  earth,  known  as  gravitation,  offers  a 
certain  resistance,  which  is  commonly  termed  weight,  to  any 
efforts  made  to  raise  the  body  from  the  surface  of  the  earth. 
Weight  is  the  characteristic  property  of  ^  all  substances  that 
can  be  handled  and  examined.     It  is  convenient  to  have  a 
collective  name  for  all  such  bodies,  and  accordingly  the  term 
matter  is  used. 

Matter,  then,  may  be  defined  as  '  *  anything  that  possesses 
weight;  that  is,  anything  that  is  acted  on  by  gravitation." 

2.  Physical  State  of  Matter. — According  to  the  tem- 
perature, matter  exists  in   one  or  another  of  three  distinct 
physical  states,  called,  respectively,  the  solid,  the  liquid,  and 
the  gaseous  state. 

A  body  in  the  solid  state  is  one  whose  molecules  change 
their  relative  position  with  great  difficulty;  as,  iron,  wood, 
silver,  steel,  etc. 

§  5 


2  THEORETICAL  CHEMISTRY.  §  5 

A  body  in  the  liquid  state  is  one  whose  molecules  change 
their  relative  position  with  ease.  Liquid  bodies  adapt  them- 
selves readily  to  the  shape  of  the  vessels  that  contain  them, 
and  their  surface  always  tends  to  become  perfectly  level. 
Water,  sulphuric  acid,  mercury,  etc.  are  liquids. 

A  body  in  the  gaseous  state  is  one  whose  molecules  tend 
to  separate  from  one  another;  as,  air,  oxygen,  hydrogen,  etc. 

Bodies  in  the  gaseous  state  are  subdivided  into  two  classes : 
permanent  gases  and  vapors. 

A  permanent  gas  is  one  that  remains  a  gas  at  ordinary 
temperatures  and  pressures. 

A  vapor  is  a  body  that  at  ordinary  temperatures  is  a  liquid 
or  solid,  but  that,  when  heat  is  applied  to  it,  becomes  a  gas, 
as  steam,  for  instance. 

One  body  may  exist  in  all  three  states ;  for  example,  mer- 
cury, which  at  ordinary  temperatures  is  a  liquid,  becomes  a 
solid  (freezes)  at  —  40°  C.  and  a  vapor  at  350°  C.  There  is 
good  reason  to  believe  that,  at  a  sufficiently  high  temperature, 
all  substances  would  be  gaseous,  and  that  at  a  sufficiently  low 
temperature  all  would  be  solid.  The  point  of  temperature 
at  which  a  solid  becomes  a  liquid  is  called  its  melting  point ; 
and  that  point  at  which,  under  ordinary  pressure,  a  liquid 
becomes  a  gas  or  vapor  is  called  its  boiling  point.  The 
number  of  units  of  heat  required  to  convert  a  unit  mass  of  a 
solid  into  a  liquid  is  called  its  heat  of  fusion,  or  liquefaction  ; 
and  the  number  of  units  of  heat  required  to  convert  a  unit 
mass  of  a  liquid  into  a  vapor,  its  heat  of  vaporization.  The 
fraction  of  a  unit  of  heat  required  to  raise  the  temperature 
of  a  unit  mass  of  any  substance  from  0°  C.  to  1°  C.  is  called 
its  specific  heat. 

3.  Division  of  Matter. — Science  assumes  three  divi- 
sions of  matter:  masses,  molecules,  and  atoms. 

A  mass  of  matter  is  any  portion  of  matter  appreciable  by 
the  senses. 

A  molecule  is  the  smallest  particle  of  matter  into  which 
a  body  can  be  divided  by  physical  means ;  it  is  the  smallest 
particle  that  is  capable  of  separate  existence. 


§5  THEORETICAL  CHEMISTRY.  3 

An  atom  is  the  still  smaller  particle  produced  by  the  divi- 
sion of  a  molecule,  and  is  regarded  by  chemists  as  the  unit 
quantity  of  chemical  combination. 

4.  Force. — It  would  appear  that  the  definition  of  matter 
given  in  Art.  1  is  sufficiently  extensive  to  embrace  every- 
thing, but  a  little  deliberation  will  show  that  there  are  other 
things  besides  matter.  To  give  an  example,  you  know  that 
a  hammer  consists  of  matter  because  it  is  heavy,  that  is, 
because  it  possesses  weight;  but  if  with  this  hammer  you 
strike  a  number  of  blows  on  a  piece  of  iron,  you  have  deliv- 
ered something  that  certainly  is  not  matter.  The  hammer 
has  not  lost  weight,  nor  has  the  iron  gained  any,  but  still 
these  blows  must  be  something,  as  they  produce  certain 
effects.  For  one  thing,  the  piece  of  iron  is  altered  in  shape, 
it  is  flattened,  and,  further,  it  has  acquired  temporary  heat. 
Again,  to  mention  another  example,  if  the  hammerhead  is 
carefully  weighed,  and  then  made  red-hot  in  a  fire,  it  will 
weigh  exactly  the  same  as  it  did  when  cold.  Further,  if  it 
is  allowed  to  cool,  this  hot  piece  of  steel  naturally  imparts 
heat  to  its  surroundings,  but  still  remains  unaltered  in 
weight.  Here,  then,  we  have  something  very  definite  that 
a  body  can  receive  and  again  yield,  and  that  certainly  is  not 
matter.  Let  us  see  what  relation  this  something  has  to 
matter.  In  the  first  example,  the  blows  were  delivered  by 
the  moving  hammer,  which  consisted  of  matter  in  motion. 
The  more  rapid  the  motion,  the  more  violent  the  blow;  in 
fact,  the  force  of  a  blow  depends  /both  on  the  quantity  of 
matter  and  on  the  rapidity  of  its  motion.  Further,  the  ham- 
mered hot  iron  and  the  heated  hammerhead  differ  from  the 
same  substances  in  their  natural  state  in  that  their  compo- 
nent particles  are  in  a  state  of  motion  (see  Art.  5,  Physics] ; 
when  these  substances  cool,  their  particles  once  more  enter 
into  a  condition  of  comparative  rest.  This  something,  then, 
that  differs  from  matter  is  closely  related  to  motion,  and  is 
known  as  force.  Force  is,  then,  that  which  is  capable  of 
setting  matter  in  motion,  or  of  changing  tJie  velocity  or  direc- 
tion of  the  motion  of  matter. 


4  THEORETICAL  CHEMISTRY.  §  5 

The  motion  of  bodies  may  be  divided  into  three  classes: 
first,  the  motion  of  the  body  as  a  whole,  as  seen  in  the  case 
of  the  moving"  hammer;  second,  the  internal  movement  of 
the  particles  of  a  body,  when  it  becomes  hot;  and  tJiird, 
atomic  movement,  which  has,  however,  not  yet  been  fully 
investigated. 

5.  Object  of  Chemistry. — We  are  now  in  a  position  to 
explain  the  objects  of  chemistry.      Matter  is  not  only  most 
varied  in  form,  but  its  form  is  also  continually  changing;  it 
is  the  office  of  the  chemist  to  investigate  these  changes,  and 
also  the  nature  of  the  substances  that  participate  in  them. 
In  short,  chemistry  is  that  science  ivliich  treats  of  the  compo- 
sition of  matter,  of  the  changes  produced  therein  by  heat  and 
other  natural  forces,  and  of  the  action  and  reaction  of  differ- 
ent kinds  of  matter  on  each  other. 

6.  Difference     Between     Chemical     and    Physical 
Changes. — It  is  of  great  importance,  even  at  this  early  stage, 
in  the  study  of  chemistry,  to  be  able  to  distinguish  between 
what  are  simply  alterations  in   the  physical   properties  of 
matter  and  what  are  chemical  changes.      A  description  of  the 
following  simple  experiments  will  greatly  facilitate  this  object : 

If  we  heat  a  piece  of  platinum  wire  in  the  flame  of  a  Bun- 
sen  burner,  we  find  that  it  soon  becomes  white  hot;  we 
remove  it  and  allow  the  wire  to  cool,  and  notice  that  it 
recovers  its  original  brightness ;  the  heat  has  not  altered  it  in 
the  least,  not  even  tarnished  it. 

Next  we  heat  in  the  same  way  a  piece  of  bright  iron  wire ; 
this  is  tarnished,  but  has  otherwise  not  undergone  any 
remarkable  change. 

Now  we  place  a  piece  of  magnesium  wire  or  ribbon  in  the 
flame ;  we  notice  at  once  a  remarkable  action,  the  wire  burns 
with  a  peculiar,  dazzling  white  light,  depositing  a  white, 
easily  powdered  body,  composed  of  oxygen  and  magnesium, 
known  as  magnesia. 

What  have  we  learned  from  these  experiments  ?  In  the 
first  case,  the  change  produced  is  merely  physical;  the  hot 
wire  possesses  properties  very  different  from  the  properties 


§  5  THEORETICAL  CHEMISTRY.  5 

it  had  when  cold,  but  as  soon  as  it  cools  again  it  regains  its 
original  character.  With  the  iron  wire  a  slight,  permanent 
change  is  produced  on  the  surface,  which  is  of  a  chemical 
character,  but  the  main  portion  of  the  wire  is  unchanged.  In 
the  third  experiment  there  has  been  a  chemical  action ;  the 
magnesium,  as  such,  has  entirely  disappeared  and  a  new  sub- 
stance has  taken  its  place. 

In  short,  we  have  learned  that  chemical  changes  are 
changes  that  take  place  within  the  molecule ;  that  they  alter 
the  character  of  the  molecule,  and  thus  cause  a  change  in  the 
identity  of  the  substance  itself. 

7.  Physical     and     Chemical    Properties. — Physical 
properties  may  be  described  as  those  properties  that  a  body 
possesses  as  a  result  of  its  molecular  conditions;  while  chem- 
ical properties  are  those  that  a  body  possesses  as  the  result 
of  the  atomic  composition  of  its  molecule. 

Tenacity,  which  signifies  the  amount  of  cohesive  attraction, 
and  color,  which  is  the  result  of  the  action  of  the  molecules 
of  a  body  upon  light,  are  properly  termed  physical  proper- 
ties, while  such  properties  as  combustibility,  explosibility, 
affinity,  etc.  are  chemical  properties. 

8.  Mechanical    Mixture    and    Chemical    Combina- 
tion.— It  is  important  to  further  distinguish  between  a  mere 
mechanical  mixture  and  a  chemical  combination;  this  dis- 
tinction is  most  easily  made  plain  by  the  study  of  an  easy, 
typical  experiment. 

If  powdered  sulphur  and  fine  copper  filings  are  mixed 
together,  the  characteristic  color  of  the  sulphur  as  well  as 
that  of  the  copper  will  disappear,  and,  to  the  unaided  eye, 
the  mixture  will  show  a  uniform  greenish  color.  By  the  aid 
of  a  microscope,  however,  the  particles  of  copper  and  sul- 
phur may  be  distinctly  seen  lying  side  by  side ;  we  can  also 
easily  separate  the  mixture  by  washing  away  the  lighter 
particles  of  sulphur,  and  at  the  same  time  restore  their  orig- 
inal colors.  Evidently  no  chemical  combination  could  have 
taken  place;  nothing  had  occurred  but  a  merely  mechanical 
mixture.  If,  however,  we  heat  this  mixture  in  a  test  tube, 


6  THEORETICAL  CHEMISTRY.  §  5 

a  remarkable  change  will  soon  take  place;  the  mixture  will 
begin  to  glow,  and  on  examining  the  mass  we  notice  that  both 
the  copper  and  the  sulphur  have  indeed  disappeared.  We 
can  no  longer  distinguish  either,  even  with  the  most  powerful 
microscope,  and  in  their  place  we  find  a  black  substance  with 
properties,  such  as  form,  appearance,  specific  gravity,  etc., 
that  are  essentially  different  from  those  possessed  by  either 
copper  or  sulphur.  Here  a  chemical  change  has  actually 
occurred.  The  copper  and  the  sulphur  under  the  influence 
of  heat  have  combined,  and  a  new  substance  with  different 
properties  has  resulted. 

These  two  experiments  show  conclusively  that  a  chemical 
combination  is  widely  different  from  a  mere  mixture.  In 
the  first  case,  the  resultant  body  is  a  mean  in  appearance, 
specific  gravity,  etc.  between  its  constituents ;  each  still  has 
its  individual  properties.  In  the  second  case,  the  resultant 
body  is  more  or  less  different  in  appearance  and  properties 
from  those  of  which  it  is  composed.  Further,  as  will  be 
demonstrated  later,  when  substances  combine  chemically  they 
invariably  do  so  in  definite  proportions.  In  a  mixture,  sub- 
stances may  evidently  be  present  in  any  proportion  whatever. 

9.  Chemical  combination  not  only  produces  bodies  differ- 
ing in  properties  from  their  constituents,  but  it  also  invariably 
produces  heat.  A  striking  example  is  the  heat  produced  in 
"slaking"  lime  (that  is,  in  adding  water  to  quicklime);  the 
heat  produced  is  entirely  due  to  the  chemical  combination  of 
water  and  quicklime. 

Hence,  in  addition  to  the  difference  in  appearance,  the 
evolution  of  heat  may  be  described  as  the  principal  charac- 
teristic feature  by  which  chemical  combination  between 
different  substances  may  be  detected. 

1C.  Indestructibility  of  Matter. — Although  chemical 
action,  such  as  combination,  certainly  can  produce  marvelous 
changes,  there  is  one  thing  that  cannot  be  accomplished :  it 
can  neither  create  nor  destroy  matter.  By  the  most  careful 
examination  of  all  known  cases  of  chemical  action,  it  has  been 
positively  proved  that  a  loss  of  matter  never  occurs,  that 


§  5  THEORETICAL  CHEMISTRY.  ? 

matter  is  indestructible,  and  that,  even  in  such  chemical 
actions  as  the  burning  and  slow  disappearing  of  a  candle, 
there  is  simply  a  change  of  state  and  not  an  annihilation  of 
matter.  The  truth  of  this  great  principle  in  chemical  science 
has  been  experimentally  demonstrated  by  showing  that  the 
weight  of  substances  acting  chemically  on  one  another  are 
always  the  same  after  chemical  changes  and  actions  have 
taken  place  as  they  were  before. 

1 1 .  Synthesis  and  Analysis. — The  composition  of  chem- 
ical compounds  is  ascertained  by  two  different  methods,  viz : 

1.  By  analysis,  which  consists  in  decomposing  the  com- 
pound into  its  constituent  elements. 

2.  By  synthesis,  which,  conversely,  consists  in  uniting  the 
constituents  and  thus  forming  the  compound. 

12.  Relation  of  Chemical  Action  to  Force. — If  some 
red  mercuric  oxide  is  placed  in  a  dry  and  clean  test  tube  and 
heated,  it  will  be  noticed  that  after  a  little  while  the  color  of 
the  mercuric  oxide  changes  from  red  to  black  and  that  the 
oxide  gradually  disappears,  while  minute  globules  of  mer- 
cury condense  on  the  upper  part  of  the  tube ;  in  short,  the 
heat  has  decomposed  the  mercuric  oxide  into  its  constituents, 
mercury  and  oxygen. 

It  has  been  stated  in  Art.  9  that  when  chemical  combina- 
tion occurs  heat  is  produced ;  in  order  to  effect  the  change  in 
the  experiment  described  above,  heat  was  used,  but  it  disap- 
peared as  heat,  being  stored  up  in  the  constituents  in  the 
form  of  the  available  or  potential  energy  of  chemical  attrac- 
tion or  affinity.  If  the  mercury  and  oxygen  are  again  com- 
bined, the  act  of  recombination  will  set  free  the  exact  amount 
of  heat  originally  necessary  to  decompose  the  compound. 
Many  instances  are  known  in  which  bodies  that  evolve  a  cer- 
tain heat  on  uniting  may  be  again  decomposed  by  applying 
a  more  intense  heat  to  them.  Thus,  the  following  rule  has 
been  established :  The  quantity  of  heat  evolved  during  chem- 
ical combination  is  exactly  equal  to  the  quantity  of  heat 
required  to  effect  the  decomposition  of  the  resultant  body. 

Further,    the    chemical   action    of   bodies   entering    into 


8  THEORETICAL  CHEMISTRY.  §  5 

combination  may  be  made  to  produce  electricity  ;  and  elec- 
tricity, again,  is  able  to  decompose  the  resultant  bodies.  There 
is  also  in  these  cases  a  definite  relation  between  the  amount 
of  chemical  action  and  the  quantity  of  electricity  produced. 

As  we  have  defined  force  as  that  which  is  capable  of  setting 
matter  in  motion,  we  must  logically  rank  chemical  combina- 
tion among  the  forces;  for  such  a  typical  chemical  action  as 
the  combustion  of  coal,  for  instance,  generates  heat,  and  thus 
indirectly,  as  in  the  steam  engine,  constitutes  our  most  valu- 
able means  of  setting  matter  in  motion. 

13.  Relation  Between  Gravitation,  Cohesion,  and 
Chemical  Attraction. — As  we  have  seen  in  Art.  4,  Physics, 
matter  is  maintained  in  its  normal  state  by  the  action  of  grav- 
itation, cohesion,  and  chemical  attraction.  Gravitation  acts 
through  all  space,  and  exerts  an  attraction  between  all  mat- 
ter. Cohesion  acts  only  across  non-appreciable  distances, 
holding  together  the  particles  of  solids,  and  also,  though  to 
a  less  extent,  those  of  liquids. 

Cohesion  may  very  often  be  overcome  by  heat ;  for  example, 
ice  may  be  converted  into  water,  and  a  further  increase  in 
the  temperature  will  destroy  the  still  prevailing  cohesion  and 
will  produce  steam,  which,  in  common  with  all  other  gases, 
is  devoid  of  cohesion  between  the  particles.  But  whether 
ice,  water,  or  steam,  the  substance  remains,  so  far  as  its 
chemical  properties  are  concerned,  exactly  the  same.  If, 
however,  we  apply  a  still  more  intense  heat,  the  steam  will 
be  decomposed  into  oxygen  and  hydrogen.  Hence,  heat  not 
only  overcomes  cohesion,  but  also  destroys  the  affinity,  or 
chemical  attraction,  that  binds  the  atoms  of  substances 
together. 

WEIGHTS    AKD    MEASURES. 


METRIC    SYSTEM  OF  WEIGHTS  AND  MEASURES. 

14.  Instead  of  employing  the  rather  complicated  English 
system  of  weights  and  measures,  most  chemists  substitute 
the  very  simple  and  convenient  metric  system, 


§  5  THEORETICAL  CHEMISTRY.  9 

The  metric  system  is  based  on  the  meter,  which,  according 
to  the  United  States  Coast  and  Geodetic  Survey  Report  of 
1884,  is  equal  to  39. 370432  inches.  The  value  commonly  used 
is  39.37  inches,  and  is  so  authorized  by  the  United  States 
Government. 

There  are  three  principal  units:  the  meter,  the  liter  (pro- 
nounced lee-ter),  and  the  gram — the  unit  of  length,  capacity, 
and  weight,  respectively.  Multiples  of  these  units  are 
obtained  by  prefixing  to  the  names  of  the  principal  units  the 
Greek  words  deka  (10),  hekto  (100),  and  kilo  (1,000) ;  the  sub- 
multiples,  or  subdivisions,  are  obtained  by  prefixing  the 
Latin  words  deci  (yV)>  centi  (TTTO)>  anc^  milli  (roVo)-  These 
prefixes  form  the  key  to  the  entire  system.  In  the  following 
tables,  the  abbreviations  of  the  principal  units  of  these  sub- 
multiples  begin  with  a  small  letter,  while  those  of  the  mul- 
tiples begin  with  a  capital  letter.  The  abbreviations  are 
those  adopted  by  the  International  Congress  of  Metric 
Weights  and  Measures.  Chemists  commonly  use  c.  c. 
instead  of  cm. 3  for  cubic  centimeter. 


TABLE   1. 
MEASURES  OF  LENGTH. 

10  millimeters  (mm.) =  1  centimeter cm. 

10  centimeters =  1  decimeter dm. 

10  decimeters =  1  meter . .  m. 

10  meters =  1  dekameter Urn. 

10  dekameters. =  1  hektometer Hm. 

10  hektometers =  1  kilometer Km. 

MEASURES  OF  SURFACE   (NOT  LAND). 

100  square  millimeters  (mm2.) =  1  square  centimeter cm2. 

100  square  centimeters —  1  square  decimeter dm2. 

100  square  decimeters =  1  square  meter m2. 

MEASURES  OF  VOLUME. 

1,000  cubic  millimeters  (mm3.)..  r.  =  1  cubic  centimeter cm3. 

1,000  cubic  centimeters =  1  cubic  decimeter dm3. 

1,000  cubic  decimeters =1  cubic  meter m3. 


10 


THEORETICAL  CHEMISTRY. 


10  milliliters  (ml.) 

10  centiliters 

10  deciliters 

10  liters 

10  dekaliters 

10  hektoliters. . , 


MEASURES  OF  CAPACITY. 

=  1  centiliter  . . . 

=  1  deciliter 

=  1  liter 

=  1  dekaliter  . . . 
=  1  hektoliter. . . 
=  1  kiloliter... 


cl. 

dl. 
1. 

Dl. 
HI. 
Kl. 


NOTE. — The  liter  is  equal  to  the  volume  occupied  by  1  cubic  decimeter 
of  pure  distilled  water  at  its  temperature  of  maximum  density,  or  4°  C. 


MEASURES  OF  WEIGHT. 


10  milligrams  (mg.). . 

10  centigrams 

10  decigrams 

10  grams 

10  dekagrams 

10  hektograms 

1,000  kilograms 


1  centigram eg. 

1  decigram dg. 

1  gram g. 

1  dekagram Dg. 

1  hektogram Hg. 

1  kilogram Kg. 

Iton..  T. 


NOTE. — The  gram  is  the  weight  of  1  cubic  centimeter  of  pure  dis- 
tilled water  at  a  temperature  of  4°  C. ;  the  kilogram  is  the  weight  of 
1  liter  of  water ;  the  ton  is  the  weight  of  1  cubic  meter  of  water. 


TABLE  2. 

CONVERSION  TABLE:   ENGLISH  MEASURES  INTO  METRIC. 


English. 

Metric. 

Metric. 

Metric. 

Metric. 

Inches  to 
Meters. 

Feet  to 
Meters. 

Pounds  to 
Kilograms. 

Gallons  to 
Liters. 

1 

0.0253998 

0.3047973 

0.4535925 

3.7853122 

2 

0.0507996 

0.6095946 

0.9071850 

7.5706244 

3 

0.0761993 

0.9143919 

1.3607775 

11.3559366 

4 

0.1015991 

1.2191892 

1.8143700 

15.1412488 

5 

0.1269989 

1.5239865 

2.2679625 

18.9265610 

6 

0.1523987 

1.8287838 

2.7215550 

22.7118732 

w 
^ 

0.1777984 

2.1335811 

3.1751475 

26.4971854 

8 

0.2031982 

2.4383784 

3.6287400 

30.2824976 

9 

0.2285980 

2.7431757 

4.0823325 

34.0678098 

10 

0.2539978 

3.0479730 

4.5359250 

37.8531225 

THEORETICAL  CHEMISTRY. 


11 


TABLE  III. 
CONVERSION  TABLE:   METRIC  MEASURES  INTO  ENGLISH. 


Metric. 

English.. 

English. 

English. 

English. 

Meters  to 
Inches. 

Meters  to 
Feet. 

Kilograms  to 
Pounds. 

Liters  to 
Gallons. 

1 

39.37043*2 

3.2808693 

2,2046223 

0.2641790 

2 

78.740864 

6.5617386 

4.4092447 

0.5283580 

3 

118.111296 

9.8426079 

6.6138670 

0.7925371 

4 

157.481728 

13.1234772 

8.8184894 

1.0567161 

5 

196.852160 

16.4043465 

11.0231117 

1.3208951 

6 

236.222592 

19.6852158 

13.2277340 

1.5850741 

7 

275.593024 

22.9660851 

15.4323564 

1.8492531 

8 

314.963456 

26.2469544 

17.6369787 

2.1134322 

9 

354.333888 

29.5278237 

19.8416011 

2.3776112 

10 

393.704320 

32.8086930 

22.0462234 

2.6417902 

15.  By  using-  these  tables,  one  measure  can  be  converted 
into  another  by  simple  addition.  It  should  be  remarked 
that  it  is  not  necessary  to  use  all  the  figures  of  the  different 
decimals,  four  decimal  places  being  all  that  are  commonly 
used;  it  is  only  in  very  exact  calculations  that  all  seven 
decimal  places  are-  required. 

ILLUSTRATION. — Change  6,471.8  feet  into  meters.  Any  number,  as 
6,471.8,  may  be  regarded  as  6,000  +  400  +  70  +  1  +  .8;  also  6,000 
=  1,000  X  6;  400  =  100  X  4,  etc.  Hence,  looking  in  the  left-hand 
column  of  Table  2  for  figure  6  (the  first  figure  of  the  given  num- 
ber) we  find  opposite  in  the  third  column,  which  is  headed  "Feet  to 
Meters,"  the  number  1.8287838.  Now,  using  but  four  decimal  places 
and  increasing  the  figure  in  the  fourth  decimal  place  by  1  (since  the 
next  figure  is  greater  than  5),  we  get  1.8288.  In  other  words,  6  feet 
=  1.8288  meters;  hence,  simply  moving  the  decimal  point  three  places 
to  the  right,  6,000  feet  =  1, 000  X  1.8288  =  1,828.8.  In  like  manner, 
400 feet  =  121.92 meters;  70 feet  =  21. 336 meters;  1  foot  =  .3048 meter: 


12  THEORETICAL  CHEMISTRY.  §  5 

and  .8  foot  =  .2438  meter.      Adding,  as  shown  below,  the  result  is 
1,972.6046  meters. 

1  828.8000 

1  2  1.9200 
21.3360 

.3048 

.2438 

1  9  7  2.6  0  4  6 

Again,  convert  19.635  Kg.  into  pounds.  The  work  should  be  perfectly 
clear  from  the  explanation  given  below.  The  result  is  43.2877  pounds. 

2  2.0  4  6  2 
19.8416 

1.3228 

.0661 

.0110 
4  3.2  8  7  7 

The  only  difficulty  in  applying  these  tables  lies  in  locating 
the  decimal  point.  Its  position  may  always  be  determined 
by  applying  the  following  statement :  If  the  figure  consid- 
ered lies  to  the  left  of  the  decimal  point,  count  each  figure  in 
order,  beginning  with  units  (but  calling  units  place  zero) 
until  the  desired  figure  is  reached ;  then  move  the  decimal 
point  to  the  right  as  many  places  as  the  figure  being  con- 
sidered is  to  the  left  of  the  decimal  point.  Thus,  in  the 
number  6,471.8  above,  the  6  lies  three  places  to  the  left  of 
the  1,  which  is  in  units  place;  hence,  the  decimal  point  is 
moved  three  places  to  the  rig/it.  By  exchanging  the  words 
"right"  and  "left,"  the  statement  will  also  apply  to  deci- 
mals. Thus,  in  the  second  case  above,  the  5  of  the  number 
19. 635  lies  three  places  to  the  rig/it  of  units  place;  hence,  the 
decimal  point  in  the  number  taken  from  the  table  is  moved 
three  places  to  the  left. 


ENGLISH  WEIGHTS  AND  MEASURES. 

10.  Familiarity  with  English  weights  and  measures  is 
assumed ;  still,  the  following  particulars  may  prove  useful : 

One  United  States  gallon  has  a  volume  of  231  cubic  inches, 
and  contains  4  quarts,  or  8  pints.  A  gallon  of  pure  water 
weighs  128  ounces,  or  58,318  grains;  hence,  a  pint  of  pure 


§  5  THEORETICAL  CHEMISTRY.  13 

water  weighs  1C  ounces,  or  1  pound,  and  from  this  has  come 
the  saying  "a  pint  is  a  pound,"  although,  of  course,  a  pint 
of  any  liquid  having  a  specific  gravity  differing  from  that  of 
water  would  have  a  different  weight.  The  measure  termed 
^  fluid  ounce  is  a  measure  of  volume,  and  not  of  weight,  and 
is  equal  to  y1^  part  of  a  pint.  In  other  words,  it  is  the  vol- 
ume of  1  ounce  of  pure  water. 

The  fluid  ounce  bears^the  same  relation  to  the  avoirdupois 
ounce  as  does  the  cubic  centimeter  to  the  gram.  When 
directions  are  given  to  dissolve  1  part  by  weight  of  a  solid 
in  10  parts  (or  any  other  number  of  parts)  by  weight  of 
water,  either  ounces  in  fluid  ounces  or  grams  in  cubic  centi- 
meters may  be  taken. 

As  it  is  the  general  custom  to  use  the  centigrade  ther- 
mometer in  all  chemical  calculations  and  investigations,  the 
temperatures  given  in  this  and  the  following  Instruction 
Papers  will  always  be  in  degrees  centigrade,  unless  other- 
wise stated. 

The  student  should  bear  in  mind  that  the  absolute  tem- 
perature of  a  body  is  its  temperature  in  degrees  C.  -f-  273-1°, 
but  in  all  practical  work  the  fraction  is  dropped  and  273°  is 
used. 


THE  BALANCE. 

17.  Though  the  student  has,  at  this  early  stage,  no  need 
of  a  balance  and  a  set  of  weights,  a  full  knowledge  of  how  to 
manipulate  the- former  is  of  such  vast  importance  to  chemists 
that  a  full  description  will  not  be  out  of  place  at  this  point. 

A  most  useful  form  of  balance,  having  a  weighing  capacity 
of  from  1  milligram  to  500  grams,  is  shown  in  Fig.  1.  The 
balance  is  enclosed  in  a  glass  case  with  sliding  doors,  and 
rests  on  a  solid  wooden  box  A  A,  in  which  is  a  drawer  that 
may  be  used  for  weights,  etc.  A  metallic  pillar  B  is  screwed 
to  the  top  of  the  box,  and  supports,  in  a  cradle  C  at  the  top, 
the  central  knife  edges  of  the  beam  D.  On  the  ends  of  the 
beam,  at  E,  are  also  fitted  knife  edges,  on  which  hang  the 
hooks  F,  and  from  these  are  suspended  the  bows  G  which 


14 


THEORETICAL  CHEMISTRY. 


carry  the  pans  H.  In  the  center  of  the  front  of  the  base  of 
the  balances  is  fixed  a  handle,  or  knob  /.  In  the  position 
shown  in  the  figure,  the  pans  are  at  rest  on  the  balance  base. 
But,  on  turning  this  knob  to  the  right,  the  whole  of  the -bal- 
ance is  raised  by  means  of  an  eccentric,  which  actuates  a 
sliding  rod  within  the  pillar  B,  and  raises  the  cradle  C.  The 


FIG.  i. 

pans  are  then  free  to  swing,  and-  their  motion  is  shown  by 
the  index  finger  K,  which  is  fastened  to  the  center  of  the 
beam.  The  end  of  this  index  moves  in  front  of  a  graduated 
scale,  which  has  a  zero  mark  in  the  center.  Before  using  the 
balance,  see  that  it  is  level  and  that  the  index  is  exactly  in 
front  of  the  0-ma.rk.  On  raising  the  pans  by  turning  the 
handle  /,  the  index  should  vibrate  an  equal  number  of 
degrees  each  side  of  the  zero.  Should  it  not  do  so,  one  of 
the  pans  is  slightly  heavier  than  the  other,  probably  through 
not  being  quite  clean ;  if  so,  turn  the  handle  back  and  set 
the  balance  to  rest  and  then  dust  the  whole  instrument  very 


THEORETICAL  CHEMISTRY. 


15 


carefully,  and  test  again.  If  still  out,  the  balance  may 
require  adjusting,  which  is  done  by  means  of  a  screw  at  the 
end  of  the  beam.  If  the  left-hand  pan  appears  to  be  too 
heavy,  the  screw  must  be  screwed  outwards  from  the  center 
of  the  beam,  and  in  the  opposite  direction  if  the  right  pan  is 
too  heavy.  However,  do  not  be  too  quick  in  adjusting;  try 
repeatedly  to  get  the  scales  to  balance  before  touching  the 
adjustment  screw;  many  a  good  balance  has  been  spoiled 

and  accurate  weighing  has 
been  made  impossible  by  too 
frequent  adjustment. 

Every  chemist  should  keep 
his  balance  and  weights  sacred, 
and  should  treat  them  with  tJie 
greatest  care  and  consideration. 
The  weights  are  usually  kept 
in  a  box,  as  shown  in  Fig.  2. 
The  larger  weights  are  gener- 
ally made  of  brass,  while  the  smaller  ones  are  made  of 
aluminum;  each  weight  rests  in  a  separate  compartment 
under  a  glass  lid  A  B.  For  the  purpose  of  lifting  the 


FIG.  3. 


weights,  a  forceps  C  is  used,  which  has  its  place  in  the  box. 
Accurate  weights  must  tinder  no  consideration  be  touched 
with  the  fingers,  but  always  lifted  with  the  forceps. 


16  THEORETICAL  CHEMISTRY.  §  5 

The  arrangement  of  the  weights  varies  with  different 
makes.  Fig.  3  shows  the  usual  and  most  convenient 
arrangement  for  a  set  of  weights  of  from  500  grams  to  .0001 
gram. 

Before  attempting  to  weigh,  the  student  must  learn  not 
only  the  denomination  of  each  weight,  but  also  its  place  in 
the  box.  He  must  be  quite  as  well  able  to  re&d  tJie  weights 
he  has  placed  in  the  balance  pan  from  the  empty  space  in  the 
box  as  from  the  weights  themselves. 

18.  Method  of  Weighing. — Let  us  suppose  that  it 
becomes  necessary  to  weigh  a  glass  flask.  This,  in  the  first 
place,  must  be  thoroughly  cleaned  and  dried,  then  placed  on 
the  left-hand  pan  of  the  balance.  Let  us  assume  the  weight 
of  the  flask  to  be  238. 847  grams  and  let  us  see  how  these 
figures  are  obtained.  First,  with  the  balance  at  rest,  place 
the  apparatus  in  position.  Then  take  the  200-gram  weight 
from  the  box  by  means  of  the  forceps  and  place  it  on  the 
right-hand  pan ;  lift  the  balance  by  turning  the  knob  /  to 
the  right,  and  notice  which  pan  is  heavier ;  lower  the  balance 
by  returning  the  knob.  In  this  case  the  weight  will  not  be 
heavy  enough,  and  the  index  will  swing  to  the  right.  Put 
in  addition  the  100-gram  weight  in  the  pan  and  lift  the 
balance — the  weight  is  too  heavy  and  the  index  will  swing  to 
the  left.  Do  not  forget  that  every  time  a  weight  (even  the 
smallest)  is  added  or  removed,  the  balance  must  be  set  to 
the  position  of  rest;  i.  e. ,  lowered  by  turning  the  handle  to 
the  left,  and  that  this  must-  be  done  very  gently  and  care- 
fully. Take  off  the  100-gram  weight  and  try  50 — too  much ; 
remove  the  50  and  try  the  20 — not  enough ;  add  another  20 
—too  much;  remove  the  20  and  add  10 — not  enough;  add  5 
—not  enough;  add  2 — not  enough;-  add  another  2 — too 
much;  remove  the  2  and  add  1 — not  enough.  As  the 
weights  get  nearer  and  nearer  to  the  weight  of  the  flask  the 
beam  begins  to  swing  more  and  more  slowly.  Having  now 
got  to  a  point  at  which  another  gram  more  than  turns  the 
scale,  commence  in  the  same  way  with  the  fractions  of  a 
gram.  Try  first  .5  gram — not  enough;  add  .2 — not  enough; 


§  5  THEORETICAL  CHEMISTRY.  17 

add  another  .2 — too  much;  remove  the  .2  and  try  .1 — not 
enough.  As  at  this  stage  the  beam  swings  very  slowly,  it  is 
advisable  to  start  it  by  wafting  a  current  of  air  down  on  one 
of  the  pans  with  the  hand,  until  the  index  finger  swings  very 
nearly  to  the  extremity  of  the  scale.  Try  next  the  .  05-gram 
weight ;  the  index  finger  swings  perhaps  just  6  divisions  to 
the  left  and  5  to  the  right — the  weight  is  too  much ;  remove 
the  .05  and  try  .02 — not  enough;  add  another  .02 — not 
enough.  Next  add  .005 — not  enough;  but  more  likely  than 
not-  the  index  finger  swings  only  a  fraction  of  a  degree 
farther  to  the  right  than  to  the  left.  Add  .002— the  index 
finger  swings  the  same  number  of  degrees  each  side  of 
zero.  The  weights  now  exactly  balance  the  flask.  Now 
we  must  read  the  weights;  this  must  first  be  done  from 
the  box,  reading  the  empty  spaces.  In  this  case  we  have 
200  +  20  +  10  +  5  +  2  4-1  =  238.  Against  the  words  "weight 
of  the  flask  "  write  this  number  in  your  note  book.  Next 
read  off  the  decigram  weight  spaces ;  there  are  empty 
.5  +  .2  +  .1  =  .8.  Write  "8"  after  238.  The  centigrams 
come  next;  there  are  .02 +  .02  =  .04.  Write '"4"  after 
238. 8.  The  milligrams  are  read  in  the  same  manner,  and 
are  .005  +.002  =  .007.  Write  "7"  after  238.84.  The  whole 
figure  will  then  read,  "Weight  of  flask  =  238.847  grams." 

Having  thus  read  the  weight  from  the  empty  spaces  in  the 
box,  next  take  out  the  weights  and  check  off  your  reading  as 
the  weights  are  returned  to  their  proper  places.  This  double 
reading  greatly  reduces  the  chances  of  error  in  recording  the 
weight  of  a  substance.  This  operation  has  been  described 
in  full,  because  it  is  the  foundation  of  exact  chemical  work; 
these  operations  are  much  shorter  in  practice  than  they 
appear  on  paper,  and  the  student  should  in  no  way  become 
discouraged. 

Remember  in  every  case  to  use  the  greatest  care 
when  working  with  a  balance;  never  touch  scales 
or  weights  with  the  fingers,  nor  leave  them  about  out 
of  their  proper  place  in  the  box;  on  the  contrary,  take 
every  precaution  to  see  that  they  are  not  soiled  or  other- 
wise injured. 


18  THEORETICAL  CHEMISTRY.  §  5 

LAWS  OF  CHEMICAL  COMBINATION. 


MOI^ECUXES    AND    ATOMS. 

19.  Chemical  Definition  of  Molecule. — A  molecule  is 
a  group  of  two  or  more   atoms  that  are   united  by  their 
affinity.      It  is  the  smallest  part  of  any  substance  that  can  be 
obtained  by  physical  division  and  still  exist  in  a  free  or 
uncombined  state. 

20.  Classification     of     Molecules.  —  Molecules     are 
divided  into  two  classes: 

1.  Elemental  molecules,  which  are  formed  by  like  atoms. 

2.  Compound  molecules,    which   are    formed   by   unlike 
atoms. 

Matter  composed  of  molecules  containing-  like  atoms  is 
termed  simple,  or  elementary,  matter  ;  matter  whose  molecules 
are  composed  of  dissimilar  atoms  is  called  compound  matter. 

ILLUSTRATION. — A  mass  of  gold,  of  silver,  of  charcoal,  of  sulphur,  etc., 
is  formed  of  molecules  whose  atoms  are  alike ;  therefore,  these  sub- 
stances are  simple,  or  elementary,  matter.  The  molecules  that  com- 
pose a  mass  of  marble,  of  sulphuric  acid,  or  of  water  are  composed  of 
unlike  atoms ;  these  substances  are  examples  of  compound  matter. 

2 1 .  The  simplest  way  to  distinguish  elemental  molecules 
from  compound  molecules  is  to  cause  a  rearrangement  of  the 
atoms  between  two  similar  molecules.    Elemental  molecules 
do  not  yield  any  new  kind  of  matter,  whereas  compound 
molecules  produce  elemental  molecules. 

Let  us,  for  instance,  assume  that  a  a  and  a  a  are  two  mole- 
cules, each  composed  of  two  atoms  a  and  a ;  then  we  never 
can  obtain,  by  any  rearrangement,  any  other  molecules  than 
a  a  and  a  a ;  but  should  the  molecules  be  a  b  and  a  b,  that  is, 
compound  molecules,  and  each  be  composed  of  the  dissimilar 
atoms  a  and  b,  then,  by  a  rearrangement  of  the  atoms,  the 
elemental  molecules  a  a  and  b  b  would  result. 

22.  dumber  of  Elemental  Molecules. — Although  the 
number  of  substances  we  observe  around  us  seems  practi- 
cally unlimited,  yet  there  are  comparatively  few  elemental 


§  5  THEORETICAL  CHEMISTRY.  19 

molecules.  The  number  of  these  that  has  been  positively 
ascertained  so  far  is  approximately  seventy,  and,  as  every 
elemental  molecule  is  composed  of  atoms  that  are  similar  to 
the  molecule,  it  is  self-evident  that  the  number  of  elemental 
atoms  is  also  approximately  seventy. 

23.  Nomenclature    of    Elemental    Molecules     and 
Atoms. — Elemental  molecules  and  their  atoms  always  pos- 
sess the  same  name,  which,  in  some  instances,  is  the  one  by 
which  these  substances  are  known  in  every-day  life,  as,  for 
instance,  gold,   silver,  iron,  copper,  etc. ;  in  other  cases  the 
name  is  chosen  on  account  of  certain   striking   properties, 
etc.  that  the  body  may  exhibit.      So,   for  instance,  chlorine 
obtained  its  name  from  the  Greek  name  of  its  color;  cczsium, 
which  is  the  Latin  name  for  "sky-blue,"  from  the  blue  lines 
which  characterize  this  element's  spectra,  etc. 

24.  Avogadro's   Law  and  the   Conclusions  There- 
from.— The  Italian  physicist  Avogadro,  in  1811,  and,  inde- 
pendently, Ampere,  a  French  chemist,  in  1814,  as  a  result  of 
various  investigations  and  experiments,  established  the  fol- 
lowing law,  which  by  right  of  precedence  is  generally  known 
as  the  law  of  Avogadro. 

Equal  volumes  of  all  substances,  cither  elemental  or  com- 
pound, in  the  gaseous  state,  at  the  same  temperature  and 
pressure,  contain  an  equal  number  of  molecules. 

From  this  law  it  obviously  follows : 

1.  That  the  molecules  of  all  gaseous  bodies  must  be  of 
equal  size. 

2.  That  the  weight  of  any  molecule — compared  with  that 
of  a  molecule  of  hydrogen — is  proportional  to  the  weight  of 
any  given  volume — also  compared  with  an  equal  volume  of 
hydrogen. 

If,  for  instance,  1  liter  of  chlorine  weighs  35.5  times  as 
much  as  1  liter  of  hydrogen,  1  molecule  of  chlorine  must 
weigh  35.5  times  as  much  as  1  molecule  of  hydrogen,  if  the 
above  law  is  true. 

25.  If   we  mix   1  volume    of  hydrogen  and    1    volume 
of  chlorine  and  expose  the  mixture  to  the  light,  we  obtain 


20  THEORETICAL  CHEMISTRY.  §  5 

2  volumes  of  hydrochloric  acid;  and  if  we  assume  that  1 
volume  of  these  gases  contained  500  molecules  of  hydrogen 
and  chlorine,  respectively,  we  will  have  1,000  molecules  of 
the  compound.  Submitting  hydrochloric  acid  to  an  analysis, 
we  find  that  each  of  its  molecules  is  composed  of  1  atom  of 
hydrogen  and  1  atom  of  chlorine;  and  since  the  1,000  mole- 
cules of  hydrochloric  acid  were  formed  from  500  molecules 
of  hydrogen  and  500  molecules  of  chlorine,  it  is  evident  that 
each  of  these  molecules  must  have  furnished  2  atoms. 

From  this  fact  we  can  state  that  a  molecule  of  hydrogen 
is  composed  of  2  atoms. 

If,  then,  we  further  assume  that  the  weight  of  an  atom  of 
hydrogen  is  1,  so  as  to  serve  as  a  unit,  the  weight  of  a  mole- 
cule of  hydrogen,  i.  e. ,  its  molecular  weight,  will  be  2. 

26.  By  density  of  a  body  is  meant  its  mass  or  quantity 
of  matter,  compared  with  the  mass  or  quantity  of  matter  of 
an  equal  volume  of  some  standard  body  arbitrarily  chosen; 
as  hydrogen  is  chosen  as  this  standard,  we  may  obtain  the 
molecular  weight  of  any  substance,  elemental  or  compound, 
by  multiplying  its  density  by  2. 

ILLUSTRATION. — :The  density  of  oxygen  is  16;  that  is,  a  given  volume 
of  it  weighs  16  times  as  much  as  the  same  volume  of  hydrogen ;  its 
molecular  weight  must,  then,  evidently  be  32,  or  16  times  greater  than 
that  of  hydrogen. 

27.  Atomicity. — The  number  of  atoms  that  compose 
any  elemental  molecule  is  obtained  by  dividing  its  molecular 
weight  by  its  atomic  weight.    We  call  an  elemental  molecule 
monatomic,    diatomic,    triatomic,    tetratomic,    or   hexatomic 
according  as  its  atomicity  is  1,  2,  3,  4,  or  6. 

Most  elemental  molecules  are  diatomic,  and  those  ele- 
ments that  are  not  volatile,  and  that,  consequently,  cannot  be 
weighed  in  the  gaseous  state,  are  arbitrarily  classed  among 
the  diatomic  molecules. 

ILLUSTRATION. — The  molecular  weight  of  chlorine  is  71,  its  atomic 
weight  is  35.5;  hence,  the  chlorine  molecule  contains  2  atoms,  or  is  said 
to  be  diatomic.  Arsenic  has  a  molecular  weight  of  300  and  an  atomic 
weight  of  75 ;  hence,  a  molecule  of  arsenic  has  an  atomicity  of  4,  or  is 
said  to  be  tetratomic. 


THEORETICAL  CHEMISTRY. 


28.     Table  4  contains  the  elemental  molecules  the  atom- 
icity of  which  has  been  experimentally  determined. 

TABLE    4. 

ATOMICITY. 


Monatomic 

Mercury 

Cadmium 

Zinc 

Barium 

Iodine  (at  1,500°) 

Bromine  (at  1,800°) 

Diatomic 
Hydrogen 
Oxygen 
Fluorine 
Chlorine 
Bromine 
Iodine  (below  1,000°) 


Nitrogen 

Sulphur  (above  800°) 

Selenium  (above  1,200°) 

Tellurium 

Tr  iatomic 
Ozone 
Selenium  (below  800°) 

Tetratomic 
Phosphorus 
Arsenic 

Hexatomic 

Sulphur  (about  500°) 


ATOMIC  WEIGHT':    VALENCE. 

29.  For  chemical  purposes  an  atom  may  be  defined  as  the 
unit  quantity  of  an  element  that  may  enter  into,  or  be  expelled 
from,  a  chemical  combination. 

The  atomic,  or  combining,  weight  is  the  relative  weight  of 
the  atom  of  each  element  compared  with  that  of  hydrogen, 
which,  being  the  lightest,  is  taken  as  unity. 

30.  Quantity  of  Combining  Power. — Assuming  the 
combining  power  of  an  atom  of  hydrogen  to  be  1,  the  com- 
bining power  of  other  atoms  will  be  1,  2,  3,  4,  5,  0,  or  7. 
This  means  that  some  atoms  possess   a   combining  power 
equal  to  that  of  hydrogen  and  unite  with  one  atom  of  it, 
while  other  atoms  possess  a  higher  combining  power  and 
unite  with  2,  3,  4,  5,  6,  or  7  atoms  of  hydrogen. 

This  quantity  of  combining  power  is  signified  by  valence; 
it  expresses  the  number  of  hydrogen  atoms  an  atom  can  com- 
bine with  or  be  exchanged  for;  since,  however,  an  atom  is 


22  THEORETICAL  CHEMISTRY.  §  5 

not  limited  to  only  one  combination  with  another  substance, 
but  can  form  several  compounds  with  it,  its  valence  neces- 
sarily varies.  When  the  valence  of  a  substance  changes,  it 
usually  increases  or  decreases  by  2  ;  so  that  an  atom  of  an 
element  may  have  a  valence  of  2,  4,  or  6,  or  of  1,  3,  5,  or  7. 
Atoms  with  even  valences  are  termed  artiads,  those  with  odd 
valences,  perissads. 

According  as  their  valence  is  1,  2,  3,  4,  5,  6,  or  7,  atoms 
are  called  monads,  dyads,  triads,  tetrads,  pentads,  Jicxads,  or 
Jieptads,  which  names  are  derived  from  the  Greek  numerals. 
When  using  the  adjective  term,  the  Latin  numerals  are  em- 
ployed, and  we  say  an  atom  is  univalent,  bivalent,  trivalent, 
quadrivalent,  quinquivalent,  sextivalent,  or  scptivalent. 

31.  To  understand  valence  fully,   we  must  remember 
that  atoms  combine  through  their  affinity.     This  affinity  is 
not  equally  strong  in  the  atoms  of  the  different  kinds  of  ele- 
mental matter.     Taking  the  affinity  of  hydrogen  as  unity, 
any  other  atom  whose  affinity  is  completely  satisfied  by  uni- 
ting with  1  atom  of  hydrogen  has  the  same  valence.     Other 
atoms  exist  whose  affinity  requires  for  saturation  2,  3,  4,  5, 
6,  or  7  atoms;  hence  their  valence  —  which  is  the  same  for  all 
similar  atoms  —  is  said  to  be  2,  3,  4,  etc.     The  valence   of 
oxygen  is  2,  because  1  atom  of  oxygen  requires  2  atoms  of 
hydrogen   to  satisfy  its  affinity.     An  atom  of   phosphorus 
demands  5  atoms  of  hydrogen  to  become  fully  saturated; 
hence  its  valence  is  5. 

32.  Graphic  Description  of  the  Valence  of  Atoms.  — 

In  order  to  have  a  tangible  idea  of  valence,  we  assume  that 
atoms  possess  a  number  of  bonds,  or  links.  This  is  graphically 


O-     O 

Monad.       Dyad.        Triad.        Tetrad.      Pentad.      Hexad.    Heptad. 

FIG.  4. 

expressed  by  a  circle  with  short  lines  radiating  from  it,  the 
number  of  which  indicates  the  valence.  The  graphic  repre- 
sentation given  in  Fig.  4  shows  the  seven  classes  of  atoms. 
The  number  of  bonds  only,  not  their  direction,  is  significant. 


THEORETICAL  CHEMISTRY. 


23 


The  valences  of  the  different  elements  are  shown  in 
Table  5.  Each  element  that  possesses  more  than  one 
valence  is  placed  under  that  heading  which  signifies  the 
most  frequently  occurring  valence  of  that  element. 

TABLE    5. 
VALENCE   OF  ATOMS.* 


PERISSADS 

ARTIADS  . 

Monads 

Dyads 

Hydrogen 
Lithium 

Oxygen 
Magnesium 

Rubidium 

Zinc 

Caesium 

Cadmium 

Potassium 

Beryllium 

Sodium 

Mercury 

Silver 

Copper 

Fluorine,  I,  III 

Calcium 

Thallium,  I,  III 

Strontium 

Chlorine,  I,  III,  V,  VII 

Barium 

Bromine,  I,  III,  V,  VII 

Cobalt,  II,  IV 

Iodine,  I,  III,  V,  VII 

Nickel,  II,  IV 

Sulphur,  II,  IV,  VI 

Selenium,  II,  IV,  VI 

Triads 
Boron 

Tellurium,  II,  IV,  VI 
Manganese,  II,  IV,  VI 

Aluminum 

Tetrads 

Gallium 

Germanium 

Indium 

Silicon 

Yttrium 

Thorium 

Scandium 

Zirconium 

Ytterbium 

Carbon 

Cerium 

Titanium,  II,  IV 

Lanthanum 

Tin,  II,  IV 

Erbium 

Platinum,  II,  IV 

Gold,  I,  III 

Palladium,  II,  IV 

Antimony,  III,  V 

Lead,  II,  IV 

Bismuth,  III,  'V 

Iron,  II,  IV,  VI 

Nitrogen,  III,  V 

Chromium,  II,  IV,  VI 

Phosphorus,  III,  V 
Arsenic,  III,  V 

Hexads 

Tungsten,  IV,  VI 

Molybdenum,  II,  IV,  VI 

Ruthenium,  II,  IV,  VI 

Pentads 

Rhodium,  II,  IV,  VI 

Columbium 

Iridium,  II,  IV,  VI 

Tantalum 

Osmium,  II,  IV,  VI 

Vanadium,  III,  V 

Uranium,  II,  IV,  VI 

*  The  valence  of  many  of  the  elements  has  not  been  determined  with 
certainty,  and  consequently  it  is  not  claimed  that  this  table  is  abso- 
lutely correct.  We  have  aimed  to  make  it  as  correct,  and  at  the  same 
time  as  simple,  as  possible. 


24  THEORETICAL  CHEMISTRY.  §  5 


SYMBOLS  A^D  FORMULAS. 

33.  For  convenience  of  description,  each  element  has  an 
abbreviation  of  its  full  name,  called  its  symbol.  These 
abbreviations  consist,  where  practicable,  of  the  initial  letter 
of  the  Latin  name  of  the  element.  As,  however,  there  are 
about  seventy  elements,  and  only  twenty-six  letters  in  the 
alphabet,  a  large  number  of  the  symbols  are  composed 
of  the  initial  and  another  distinctive  letter  selected  from 
the  name.  Thus,  the  three  elements  carbon,  chlorine, 
and  copper  (cuprum)  all  have  names  commencing  with 
UC";  carbon  being  the  most  important  has  the  letter  C 
for  its  symbol,  while  67  and  Cu  stand  for  chlorine  and 
copper,  respectively. 

As  all  compound  bodies  are  the  result  of  the  combination 
of  elements,  they  may  conveniently  be  expressed  symbolically 
by  placing  side  by  side  the  symbols  of  the  constituent  ele- 
ments. The  symbol  of  a  compound  is  termed  -a,  formula. 
Thus,  common  salt  consists  of  sodium,  whose  symbol  is  Na, 
and  chlorine,  whose  symbol  is  67;  accordingly,  NaClvs,  writ- 
ten for  its  formula. 

The  multiplication  of  atoms  is  expressed  by  placing  an 
Arabic  numeral  to  the  right  of  and  below  the  symbol. 

Thus,  S9  stands  for  2  atoms  of  sulphur;  N3  for  3  atoms  of 
hydrogen. 

The  multiplication  of  a  molecule  (formula)  is  expressed 
either  by  enclosing  the  formula  in  brackets  and  placing  the 
multiplier  outside  to  the  right  of  and  below  the  formula, 
or  by  placing  the  multiplier  to  the  left  of  and  in  front  of  the 
formula. 

Either  (H^O)^  or  4//26>  stands  for  4  molecules  of  water  ; 
either  (6>2)4  or  46>2  stands  for  4  molecules  of  oxygen,  each 
molecule  consisting  of  2  atoms. 

The  valence  of  an  atom  is  expressed  by  placing  a  Roman 
numeral  or  a  corresponding  number  of  dashes  to  the  right 
of  and  above  it.  £7iv,  or  C'": ',  stands,  for  instance,  for  the 
quadrivalent  atom  of  carbon  and  0",  or  O" ,  stands  for  the 
bivalent  atom  of  oxygen. 


§  5  THEORETICAL  CHEMISTRY.  25 

34.     Combination     in     Definite    Proportions.  —  The 

symbols,  however,  signify  not  only  the  names  of  elements, 
but  according  to  the  law  —  the  proportions  by  weight, 
according  to  which  bodies  combine,  are  invariable  for 
each  combination — these  symbols  also  indicate  the  mini- 
mum proportion  in  which  the  different  atoms  combine. 
For  instance,  O  not  only  signifies  oxygen,  but  it  also 
always  stands  for  16  parts  by  weight  of  it;  Cl,  which 
stands  for  chlorine,  always  signifies  35.5  parts  by  weight. 
In  this  way  each  of  the  elements  has  its  own  symbol, 
representing  the  name  and  the  definite  proportion,  by 
weight,  in  which  it  combines. 

From  this  it  is  evident  that  we  not  only  express  by  a  for- 
mula the  qualitative,  but  also  the  quantitative,  composition 
of  a  substance. 

ILLUSTRATION. — H^SO*  stands  for  sulphuric  acid,  consisting  of 
//a  =  2  parts  by  weight  of  hydrogen ; 
S   =  32  parts  by  weight  of  sulphur ; 
(9  4  =  64  parts  by  weight  of  oxygen. 


35.     Combination     in     Multiple    Proportions. — The 

combinations  of  oxygen  and  hydrogen,  H9O  (water)  and 
Hfli  (hydrogen  peroxide),  afford  examples  of  two  elements 
combining  in  more  than  one  proportion.  It  will  be  observed 
from  the  two  formulas  that  in  the  latter  compound  exactly 
twice'  the  quantity  of  oxygen  is  combined  with  the  same 
quantity  of  hydrogen  as  in  the  former.  Taking,  also, 
other  series  of  compounds  into  consideration,  it  will  be 
seen  that  a  simple  relation  exists  between  the  propor- 
tions of  each  element  dn  the  various  compounds.  In 
fact,  not  only  does  chemical  combination  invariably  occur 
in  definite  proportion,  but  when  tivo  elements  combine  in 
more  than  one  proportion  they  unite  in  multiple  propor- 
tions. The  reason  why  combination  occurs  in  multiple 
proportion  is,  that  when  two  elements  combine,  1  atom 
of  the  first  may  combine  with  either  1,  2,  or  3  atoms 
of  the  second,  but  combination  with  the  fraction  of  an 


26  THEORETICAL  CHEMISTRY.  §  5 

atom   would  necessarily   be   a   contradiction  of  the  atomic 
theory. 

ILLUSTRATION. — Nitrogen  forms  five  distinct  compounds  with  oxygen ; 
if  such  quantities  of  the  compound  be  taken  as  contain  the  same  weight 
of  nitrogen,  the  weights  of  oxygen  will  be  proportional  to  the  numbers 
1,  2,  3,  4,  and  5. 

Hyponitrous  oxide  A^O,  28  parts  of  nitrogen  and  16  parts  of  oxygen. 

Nitrogen  dioxide   Ar2<92,  28  parts  of  nitrogen  and  32  parts  of  oxygen. 

Nitrous  oxide          A^Og,  28  parts  of  nitrogen  and  48  parts  of  oxygen. 

Nitrogen  tetroxide  N9O4,  28  parts  of  nitrogen  and  64  parts  of  oxygen. 

Nitric  oxide  N2O5,  28  parts  of  nitrogen  and  80  parts  of  oxygen. 


THE  CHEMICAL  ELEMENTS. 

36.  The  names  of  the  elements,  their  symbols,  and  their 
combining  weight   are  given  in  Table  6.     The  combining 
weights   given   are   those   derived  from  experiments.     For 
most  calculations  the  nearest  whole  number  or  whole  num- 
ber with  .5  may  be  employed.     Thus,  chlorine  maybe  taken 
as  35.5,  oxygen  as  16,  and  so  on. 

37.  Metals  and  !Non-Metals. — The  elements  have  been 
divided  into  two  groups,    according   to   whether   they   are 
metallic  or  non-metallic  in  their  nature.     The  metallic  prop- 
erties are  very  decided  in  a  metal  such  as  either  iron,  copper, 
or  gold;  and  it  is  equally  obvious  that  such  bodies  as  sulphur 
and  phosphorus  are  not  metals.     But,  although  the  proper- 
ties of  these  are  very  definite,  a  number  of  other  bodies  are 
much  less  marked  in  character  ;  in  fact,   there  is  no  well 
denned  line  of  division  between  the  two  classes,  as  the  one 
vseries   gradually   overlaps    the   other.     Thus,    the   element 
arsenic,  which  occupies  an  intermediate  position,  is  placed  by 
some  chemists  among  the  metals,  and  by  others  among  the 
non-metals  or  metalloids. 

In  physical  character  the  metals  are,  as  a  rule,  opaque  bodies 
with  a  peculiar  luster,  known  as  metallic  luster,  and  they  are 
comparatively  good  conductors  of  heat  and  electricity.  These 


§  5  THEORETICAL  CHEMISTRY.  27 

properties,  however,  do  not  belong  exclusively  to  the  metals, 
for  carbon  in  the  form  of  graphite  has,  as  is  well  known, 
a  very  decided  metallic  luster,  and  conducts  electricity 
well. 

Chemically,  the  metals,  as  a  whole,  form  oxides  that  act  as 
bases,  while  the  non-metallic  oxides  form  acids,  but  even  in 
this  respect  the  series  overlap  each  other.  (The  meaning  of 
the  terms  oxide,  base,  and  acid  is  fully  explained  in  subse- 
quent articles.)  The  non-metals  in  Table  6  are  indicated 
by  being  printed  in  heavy  type,  while  the  metals  are  printed 
in  ordinary  type. 

As  hydrogen  is  the  lightest  known  element,  it  has  been 
customary  to  call  its  atomic  weight  1,  and  state  the  atomic 
weights  of  other  elements  in  terms  of  hydrogen  as  1.  That 
is,  the  atomic  weight  of  an  element  has  meant  that  the  ele- 
ment was  that  many  times  as  heavy  as  hydrogen.  As  it 
somewhat  simplifies  accurate  chemical  calculations  to  call 
the  atomic  weight  of  oxygen  16,  there  is  a  tendency  at  present 
to  arbitrarily  assign  this  value  to  oxygen,  and  to  assign 
atomic  weights  to  the  other  elements  that  will  agree  with 
this  arbitrarily  chosen  standard. 

In  the  accompanying  table,  taken  from  the  atomic  weights 
published  early  in  the  year  1899  by  the  American  Chemical 
Society,  weights  based  on  each  of  these  standards  are  given 
in  separate  columns.  As  it  seems  more  rational  to  call  the 
lightest  element  1,  and  use  this  as  the  measure  for  the 
others,  this  standard  has  been  retained  in  this  and  the 
following  sections. 

These  atomic  weights  are  the  results  of  the  latest  deter- 
minations, and  are  probably  more  accurate  than  any  pre- 
viously published;  but  as  they  are  not  absolutely  correct, 
and  will  probably  be  changed  from  time  to  time,  it  is  thought 
best  to  use  the  atomic  weights  most  generally  used,  in 
this  and  the  succeeding  papers,  but  if  the  student  prefers 
he  can  easily  substitute  the  values  here  given  in  his  later 
work. 


THEORETICAL  CHEMISTRY. 


TABLE  6. 

NAMES  OF  ELEMENTS,  THEIR  SYMBOLS,  AND  ATOMIC 
WEIGHTS. 


Name. 

Symbol. 

Atomic  Weight. 

H  =  1. 

O  —  16. 

Aluminum   

Al 

Sb 
As 
Ba 
Bi 
B 
Br 
Cd 
Cs 
Ca 
C 
Cc 
Cl 
Cr 
Co 
Cb 
Cu 
E 
F 
G 
Ge 
Au 
H 
In 
I 
Ir 
Fe 
La 
Pb 
Li 
Mg 
Mn 

26.91 
119.52 
74.44 
136.39 
206  .  54 
10.86 
79.34 
111.54 
131.89 
39.76 
11.91 
138.30 
35.18 
51.74 
58.55 
93.02 
63.12 
165.06 
18.91 
69.38 
71.93 
195.74 
1.00 
112.99 
125.89 
191.66 
55.60 
137.59 
205.36 
6.97 
24.10 
54.57 

27.11 
120.43 
75.01 
137.43 
208.11 
10.95 
79.95 
112.38 
132.89 
40.07 
12.00 
139.35 
35.45 
52.14 
58.99 
93.73 
63.60 
166.32 
19.06 
69.91 
72.48 
197.23 
1.008 
113.85 
126.85 
193.12 
56.02 
138.64 
206.92 
7.03 
24.28 
54.99 

Antimony    (stibium)  .  .  . 
Arsenic          

Barium  

Bismuth      ....         .... 

Boron 

Bromine?  

Cadmium 

Caesium  

Calcium        

Cartoon. 

Cerium      

Chlorine 

Chromium  

Cobalt  

Columbium  

Copper  (cuprum) 

Erbium  

Fluorine 

Gallium    

Germanium 

Gold  (aurum)  

Hydro  °ren 

Indium   

Iodine 

Iridium 

Iron  (ferrum)  

Lanthanum            

Lead  (plumbum)        .  . 

Lithium  

Magnesium  

Manganese  . 

THEORETICAL  CHEMISTRY. 


29 


TABL.E  6 — Continued. 


Atomic 

Weight. 

H  =  1. 

O  =  16. 

Mercury  (hydrargyrum) 
Molybdenum 

Hg 

Mo 

198.49 
95  26 

200.00 
95  99 

Nickel  

Ni 

58  24 

58.69 

N 

13  93 

14  04 

Osmium 

Os 

189  55 

190  99 

o 

15  88 

16  00 

Palladium  

Pd 

105  56 

106  36 

Phosphorus       . 

P 

30  79 

31  02 

Platinum 

Pt 

193  41 

194  89 

Potassium  

K 

38  82 

39.11 

Rhodium 

Rh 

102  23 

103  01 

Rubidium  

Rb 

84.78 

85  .  43 

Ruthenium 

Rn 

100  91 

101   68 

Samarium 

Sin 

149  13 

150  26 

Scandium     

Sc 

43  78 

44.12 

Selenium 

St? 

78  58 

79.17 

Silicon  

Si 

28.18 

28.40 

Silver  (arp'entum) 

107  11 

107.92 

Sodium  (natrium)  

Na 

22.88 

23.05 

Strontium  

Sr 

86.95 

87.61 

Sulphur 

s 

31  83 

32.07 

Tantalum     

Ta 

181.45 

182.84 

Tellurium          .         ... 

Te 

126  52 

127.49 

Thallium  ....-...:-  

Tl 

202.61 

204.16 

Thorium         

Th 

230.87 

232  .  63 

Tin  (stannum) 

Sn 

118  15 

119.05 

Titanium  

Ti 

47.79 

48.15 

Tungsten  (wolfram)  .  .  . 
Uranium 

W 

u 

183.43 
237  77 

184.83 
239  59 

Vanadium  

V 

50.99 

51  .  38 

Ytterbium 

Yb 

171.88 

173.19 

Yttrium 

Y 

88  35 

89  02 

Zinc  

Zn 

64.91 

65.41 

Zirconium 

Zr 

89  72 

90  .  40 

30  THEORETICAL  CHEMISTRY.  §  5 

At  ordinary  temperatures,  two  of  the  elements — mercury 
and  bromine — are  liquids;  five — ox)^gen,  hydrogen,  chlorine, 
fluorine,  and  nitrogen — are  gases.  The  remaining  elements 
are  solids ;  but  at  varying  temperatures  all  have  been  lique- 
fied, with  the  exception  of  carbon,  which  so  far  has  only 
been  slightly  softened  with  the  highest  temperature  obtain- 
able. 

38.  Electropositive  and  Electronegative  Elements. 

A  more  scientific,  but  not  generally  accepted,  division  is  to 
distinguish  the  elements  according  to  the  quality  of  their 
combining  power.  This  means  that  they  are  divided  into: 

1.  Positive  elements,  or  those  that  are  attracted  to  the 
negative  electrode  in  electrolysis. 

2.  Negative  elements,  or  those  that  are  attracted  to  the 
positive  electrode. 

ILLUSTRATION. — If  salt  is  decomposed  through  the  influence  of  elec- 
tricity, it  will  be  noticed  that  the  chlorine  atoms  gather  at  the  positive 
electrode,  and  are,  consequently,  called  negative,  while  the  sodium 
atoms  collect  at  the  negative  electrode  and  are  considered  positive 
atoms. 

In  Table  7  the  principal  elements  are  arranged  according 
to  their  electrochemical  character,  each  atom  being  positive 
to  any  atom  placed  above  it,  and  negative  to  any  one  placed 
below  it. 

It  should  be  remarked  here,  that  it  is  absolutely  necessary 
that  the  student  should  make  himself  thoroughly  familiar 
with  the  names  of  the  elements,  their  symbols,  atomic 
weights,  and  electrochemical  characters;  in  fact,  he  should 
try  to  learn  these  things  by  heart ;  nothing  looks  so  bad  as  to 
see  a  chemist  referring  to  his  books  to  find  the  symbol,  com- 
bining weight,  etc.  of  an  element. 

39.  Meta-Elements. — In  Art.  22  the  number  of  ele- 
ments is  given  as  "  approximately  seventy";  the  reason  for 
not  giving  a  more  positive  figure  is  that  the  standard  adopted 
for  elements  has  become  doubtful  through  recent  researches. 

It  has,  for  instance,  been  observed  that  the  spectrum  of 


THEORETICAL  CHEMISTRY. 


31 


TABLE   7. 

ELECTROCHEMICAL  CHARACTER  OF  ELEMENTS. 


Negative  End 
Oxygen 
Sulphur 
Nitrogen 
Fluorine 
Chlorine 
Bromine 
Iodine 
Selenium 
Phosphorus 
Arsenic 
Chromium 
Vanadium 
Molybdenum 
Tungsten 
Boron 
Carbon 
Antimony 
Tellurium 
Tantalum 
Columbium 
Titanium 
Silicon 
Tin 

Hydrogen 
Gold 
Osmium 
Iridium 
Platinum 
Rhodium 
Ruthenium 
Palladium 
Mercury 
Silver 


Copper 
Uranium 
Bismuth 
Gallium 
Indium 
Germanium 
Lead 
Cadmium 
Thallium 
Cobalt 
Nickel 
Iron 
Zinc 

Manganese 
Lanthanum 
Cerium 
Thorium 
Zirconium. 
Aluminum 
Scandium 
Erbium 
Yttrium 
Ytterbium 
Beryllium 
Magnesium 
Calcium 
Strontium 
Barium 
Lithium 
Sodium 
Potassium 
Rubidium 
Caesium 
Positive  End 


32  THEORETICAL  CHEMISTRY.  §  5 

didymium  is  capable  of  resolution  into  nine  separate  constit- 
uents, and  that  yttrium  may  be  made  to  yield  five  different 
components,  each  giving  a  different  phosphorescent  spec- 
trum. From  this  the  conclusion  has  been  drawn  that  such 
elements  as  didymium,  yttrium,  samarium,  decipium,  philip- 
pium,  holmium,  thulium,  dyprosium,  terbium,  gnomium, 
etc.  are  not  elements  in  the  true  sense  of  the  word,  but  that 
they  are  formed  of  certain  simpler  substances,  which  are  at 
present  unknown  to  us,  and  which  are  provisionally  called 
meta-elements. 

4O.  Determination  of  Atomic  Weights. — The  atomic 
weight  has  been  defined  in  Art.  29  as  the  relative  weight  of 
any  atom  referred  to  hydrogen  as  unity.  In  order  to  deter- 
mine the  atomic  weight  of  an  atom,  we  have  to  know: 

1.  The  quantity,  by  weight,  of  the  element  that  combines 
with  1  atom  of  hydrogen. 

2.  The  molecular  weight  of  the  hydrogen  compound. 

As  the  absolute  weight  of  each  constituent  of  a  certain 
quantity,  usually  100  parts,  of  a  hydrogen  compound  may  be 
obtained  by  analysis;  the  quantity,  by  weight,  of  a  body  that 
unites  with  1  part  of  hydrogen  may  be  obtained  by  simple 
proportion,  and  as  the  molecular  weight  of  any  substance 
represents  the  sum  of  the  weights  of  its  constituent  atoms, 
the  combining  weights  obtained  by  analysis,  when  added 
together,  must  either  give  the  molecular  weight,  or  a  num- 
ber of  which  the  molecular  weight  is  a  multiple. 

EXAMPLE  1. — A  quantity  of  hydrogen  iodide  contains,  according  to 
analysis,  99.22  per  cent,  of  iodine  and  .78  per  cent,  of  hydrogen;  the 
molecular  weight  of  hydrogen  iodide  is  128.  What  is  the  atomic 
weight  of  iodine  ? 

SOLUTION. — The  proportion  .78  :  99.22  =  1  :  127  shows  that  in  this 
compound  the  smallest  quantity  of  iodine  that  can  combine  with  1  part 
of  hydrogen  weighs  127  times  as  much  as  hydrogen,  and  as  the  sum  of 
the  combining  weights  (127  +  1)  is  128,  which  is  also  the  molecular 
weight  of  hydrogen  iodide,  the  atomic  weight  of  iodine  is  127.  Ans. 

EXAMPLE  2. — Hydrogen  arsenide,  or  arsine,  is  shown  by  analysis  to 
contain  3.85  per  cent,  of  hydrogen  and  96.15  per  cent,  of  arsenic;  its 
molecular  weight  is  78.  What  is  the  atomic  weight  of  arsenic  ? 


§  5  THEORETICAL  CHEMISTRY.  33 

SOLUTION. — From  the  proportion  3.85  :  96.15  =  1  :  25,  it  is  seen  that 
25  parts  of  arsenic  combine  with  1  part  of  hydrogen.  If  there  is  only  1 
atom  of  hydrogen  in  1  molecule  of  arsine,  then  25,  being  the  smallest 
quantity,  by  weight,  in  which  arsenic  combines,  will  be  its  atomic 
weight ;  but  adding  the  two  combining  weights  (25  + 1)  together  we 
obtain  only  26,  which,  however,  is  exactly  one-third  of  the  given 
molecular  weight  of  arsine.  Hence,  a  molecule  of  arsine  is  composed  of 
3  parts  of  hydrogen  and  25  X  3  =  75  parts  of  arsenic.  75  is,  therefore, 
the  atomic  weight  of  arsenic.  Ans. 

41.  Indirect  Method  for  Determination  of  Atomic 
Weights. — Some  elements  do  not  unite  directly  with  hydro- 
gen ;  the  comparison  with  hydrogen  must,  then,  be  made  by 
means  of  another  element,  for  which  purpose  chlorine  is 
generally  chosen. 

EXAMPLE. — There  is  no  known  hydrogen  compound  of  silver,  but 
there  exists  a  well  known  combination  of  silver  and  chlorine,  called 
horn-silver.  On  analysis,  horn-silver  yields  75.26  per  cent,  of  silver  and 
24.74  per  cent,  of  chlorine  ;  its  molecular  weight  is  143.5.  What  is  the 
atomic  weight  of  silver  ? 

SOLUTION. — As  35. 5  parts  of  chlorine  combine  with  1  part  of  hydrogen, 
the  quantity  of  silver  that  combines  with  35.5  parts  of  chlorine  is  its 
atomic  weight';  24.74  :  75.26  =  35.5  :  108.  Since  108  +  35.5  =  143.5,  the 
molecular  weight,  108,  thus  obtained  must  be  the  atomic  weight  of 
silver.  Ans.  

EXAMPLES  FOB  PRACTICE. 

42.  Solve  the  following : 

1.  The  density  of  arsenic  vapor  is  150;    what  is  its  molecular 
weight?  Ans.  300. 

2.  The  density  of  phosphorus  is  62 ;  what  is  its  molecular  weight  ? 

Ans.  124. 

3.  The  density  of  phosphorus  is  62,  and  its  atomic  weight  is  31 ; 
how  many  atoms  compose  a  phosphorus  molecule  ?  Ans.  4  atoms. 

4.  The  density  of  bromine  is  80,  its  atomic  weight  is  80 ;  how  many 
atoms  compose  a  bromine  molecule  ?  Ans.  2  atoms. 

5.  Water,  a  compound  of  hydrogen  and  oxygen,  has  a  molecular 
weight  of  18.     On  analysis  it  yields  88.89  per  cent,  of  oxygen  and  11.11 
per  cent,  of  hydrogen ;  what  is  the  atomic  weight  of  oxygen  ? 

Ans.  16. 

6.  Marsh  gas  is  a  compound  containing  75  per  cent,  of  carbon  and 
25  per  cent,  of  hydrogen.     Its  density  is  8 ;  what  is  the  atomic  weight 
of  carbon  ?  Ans.    12. 


34  THEORETICAL  CHEMISTRY.  §  5 

7.  The  analysis  of  aurous  chloride  gives  84.7  per  cent,  of  gold  and 
15.3  per  cent,  of  chlorine.     The  density  of  the  compound  is  116 ;  what  is 
the  atomic  weight  of  gold  ?  Ans.  196.5. 

8.  The  analysis  of  stannous  chloride  gives  62.48  per  cent,  of  tin  and 
37.57  per  cent,  of  chlorine.     The  molecular  weight  of  the  compound  is 
189;  what  is  the  atomic  weight  of  tin  ?  Ans.  118. 


43.     Relation  of  Atomic  Weight  to  Specific  Heat.— 

When  equal  weights  of  different  bodies  are  raised  through 
the  same  number  of  degrees  of  temperature,  they  take  up 
different  amounts  of  heat ;  or,  in  other  words,  different  bodies 
possess  different  capacities  for  heat.  Thus,  1  kilogram  of 
water  requires  thirty  times  as  much  heat  as  1  kilogram  of 
mercury  to  raise  its  temperature  1  degree ;  that  is,  if  the 
quantity  of  heat  required  to  raise  the  temperature  of  1  kilo- 
gram of  water  1  degree  is  represented  by  1,  the  quantity  of 
heat  required  to  raise  the  same  weight  of  mercury  1  degree 
will  be  represented  by  -£$.  This  fraction  expresses  what  is 
called  the  specific  heat  of  mercury. 

The  specific  heat  of  a  solid  or  liquid  body  is,  then,  the 
amount  of  heat  required  to  raise  the  temperature  of  a  certain 
weight  of  this  body  1  degree,  the  amount  of  heat  required  to 
raise  the  temperature  of  an  equal  weight  of  water  1  degree 
being  taken  as  unity. 

The  specific  heat  of  a  substance  varies  according  as  the 
substance  is  in  the  solid,  liquid,  or  gaseous  state,  but  the 
specific  heat  of  all  solid  elements  exhibits  a  remarkable  rela- 
tion to  their  atomic  weight.  In  1819  Dulong  and  Petit 
observed  that  if,  instead  of  calculating  the  specific  heats  for 
equal  weights,  such  weights  be  taken  as  represent  the  atomic 
weights  of  the  solid  elements,  the  numbers  expressing  the 
capacity  for  heat  of  the  different  atoms  are  sensibly  equal ; 
from  which  it  follows  that  all  solid  elements  possess  the  same 
atomic  heat,  and  that  the  specific  heats  of  solid  elements  are 
inversely  as  their  atomic  weights. 

Hence,  we  have  in  the  determination  of  specific  heat  a 
means  of  checking  the  atomic  weight  of  a  solid  element,  or 
of  ascertaining  it  in  doubtful  cases.  This  is  done  by  dividing 


§  5  THEORETICAL  CHEMISTRY.  35 

the  atomic  heat  of  a  solid  element  by  its  specific  heat,  the 
quotient  thus  obtained  being  the  atomic  weight.  The  num- 
ber 6.4  represents,  approximately,  the  atomic  heat  of  the 
solid  elements. 

SPECIFIC  HEAT.     ATOMIC  HEAT.  ATOMIC  WEIGHT. 

ILLUSTRATION.— Lead .031  6.41  206.77 

Platinum  .032  6.33  197.81 

Silver...  .059  6.37  107.97 

Sulphur.   .2026  6.48  31.98 

Tin 054  6.37  117.96 

Copper..   .0951  6.04  63.51 


COMPOUND  MOLECULES. 

44.  A  compound  molecule  may  be  defined  as  a  molecule 
composed  of  dissimilar  atoms,  which  are  united  according  to 
the  law  of  valence.     The  number  of  atoms  that  take  part  in 
this   union   is   apparently  unlimited;    while  a  molecule    of 
hydrochloric  acid  contains  but  two  atoms,  and  a  molecule  of 
sulphuric  acid  7  atoms,  certain  organic  compounds  contain  a 
much  higher  number.      Protagon,  for  instance,  a  compound 
of  lecithin  and  cerebrin,  both  obtained  from  the  substance  of 
the  human  brain,  contains  384  atoms. 

The  molecular  weight  of  a  compound  molecule  is  equal  to 
the  sum  of  the  weights  of  its  component  atoms,  and  also  to 
twice  its  density  in  the  gaseous  state.  Thus,  an  atom  of 
chlorine  weighing  35.5  unites  with  an  atom  of  hydrogen 
weighing  1,  to  form  a  molecule  of  hydrogen  chloride  (hydro- 
chloric acid),  whose  molecular  weight  is  36.5.  Or,  2  atoms 
of  hydrogen  weighing  2,  1  atom  of  sulphur  weighing  32,  and 
4  atoms  of  oxygen  weighing  64  unite  to  form  a  sulphuric-acid 
molecule  weighing  OS. 

45.  Compound  molecules  are  classified  as: 

1.  Binary  molecules;  that  is,  molecules  which,  whatever 
the  number  of  atoms  maybe,  contain  only  two  different  kinds 
of  atoms. 

2.  Ternary  molecules;  that  is,  molecules  whose  dissimilar 
atoms  are  united  by  the  aid  of  some  third  atom. 


36  THEORETICAL  CHEMISTRY.  §  5 

BINARY  COMPOUNDS. 

46.  Nomenclature  of  Binary  Molecules. — The  names 
of  chemical  compounds  are  fixed  by  rule,  but  in  the  case  of 
the  more  well  known  compounds  the  old,  every-day  names 
are  commonly  used;  hardly  any  one,   for  instance,  would 
think  of  saying  ' '  hydrogen  monoxide  "  for  water.     The  name 
of  a  compound  should,  as  far  as  possible,  indicate  its  compo- 
sition; this  end  is  attained  by  making  the  name  of  a  binary 
compound  consist  of  derivatives  of  the  names  of  its  compo- 
nents.    There  are,  unfortunately,   several  modifications  of 
each  name,  and  English-speaking  chemists  have  not  always 
used  a  uniform  method  of  nomenclature ;  the  student,  there- 
fore, must  make  himself  familiar  with  all  of  them,  as,  in  the 
course  of  reading  various  books  and  manuals  on  chemistry, 
he  is  sure  to  meet  with  the  same  body  under  different  names. 
Very  little  practice  is  sufficient,  however,  to  overcome  this 
little  difficulty. 

The  names  of  the  binary  molecules  are  derived'  from  their 
constituent  atoms.  The  most  frequently  occurring  binaries 
are  those  composed  of  a  metal  and  a  metalloid  (non-metal) ; 
the  name  of  the  metal  is  first  written,  and  then  that  of  the 
metalloid,  one  or  more  syllables  being  removed  from  the 
latter  and  the  termination  ide  added. 

ILLUSTRATION. — Copper  and  oxygen  yield  copper  oxide. 

Magnesium  and  oxygen  yield  magnesium  oxide. 
Silver  and  sulphur  yield  silver  sulphide. 
Zinc  and  phosphorus  yield  zinc  phosphide. 
Calcium  and  iodine  yield  calcium  iodide. 
Aluminum  and  bromine  yield  aluminum  bromide. 
Sodium  and  chlorine  yield  sodium  chloride. 
Potassium  and  nitrogen  yield  potassium  nitride. 
Barium  and  fluorine  yield  barium  fluoride.  - 
Cadmium  and  selenium  yield  cadmium  selenide. 

The  termination  ide  is  always  characteristic  of  a  binary 
compound. 

47.  Modification    of    This    Rule. — As   it   frequently 
happens  that  more  than  one  compound  of  the  same  elements 


§  5  THEORETICAL  CHEMISTRY.  37 

is  known,  it  becomes  necessary  to  distinguish  these  com- 
pounds. 

Whenever  the  metal  atom  enters  into  combination  with 
more  than  one  valence,  this  fact  is  indicated  in  the  com- 
pound by  changing  the  termination  of  the  name  of  this  atom 
into  ic  or  ous. 

Should  the  metal  act  with  only  two  valences,  then  in 
the  higher  one  its  name  takes  the  termination  ic,  and  in  the 
lower  one  the  termination  o us. 

ILLUSTRATION. — Bivalent  mercury  and  oxygen  yield  murcuric  oxide. 
Univalent   mercury   and  oxygen    yield    mercurous 

oxide. 

Quadrivalent  tin  and  chlorine  yield  stannic  chloride. 
Bivalent  tin  and  chlorine  yield  stannous  chloride. 

48.  Use  of  Prefixes. — If  the  metallic  constituent  acts 
with  more  than  two  valences,  the  termination  ic — being 
given  on  the  discovery  of  the  compound — is  generally  arbi- 
trarily assigned,  and  a  further  discrimination  becomes  again 
necessary;  this  distinction  is  marked  by  the  use  of  a  prefix. 
A  compound  in  which  the  valence  of  the  first,  or  metallic, 
constituent  is  less  than  in  the  ous,  takes  the  prefix  Jiypo, 
which  means  "under."  When  the  valence  is  above  ic,  the 
prefix  per  is  used.  The  termination  of  the  second  con- 
stituent, however,  always  remains  ide. 

ILLUSTRATION. — Quinquivalent    chlorine   and  oxygen   form   chloric 

oxide. 

Trivalent  chlorine  and  oxygen  form  chlorous  oxide. 
Univalent  chlorine  and  oxygen  form  hypochlorous 

oxide. 
Septivalent   chlorine   and    oxygen   form  perchloric 

oxide. 

Sexivalent  sulphur  and  oxygen  form  sulphuric  oxide. 
Quadrivalent  sulphur  and  oxygen  form  sulphurous 

oxide. 
Bivalent  sulphur  and  oxygen  form  hyposulphurous 

oxide. 

Quinquivalent  nitrogen  and  oxygen  form  nitric  oxide. 
Trivalent  nitrogen  and  oxygen  form  nitrous  oxide. 
Univalent  nitrogen   and  oxygen   form  hyponitrous 

oxide. 


38  THEORETICAL  CHEMISTRY.  §  5 

49.  Use   of  ^Numeral  Prefixes. — In  order  to  indicate 
the  number  of  atoms  of  each  constituent,  the  Greek  numer- 
als are  prefixed  to  the  name  of  the  constituent. 

ILLUSTRATION. — 1  atom  of  Cand  2  atoms  of  O  form  carbon  dioxide. 

1  atom  of  P  and  5  atoms  of  Br  form  phosphorous 
pentabromide. 

2  atoms  of  Fe  and  3  atoms  of  O  form  diferric  trioxide. 
8  atoms  of  Ti  and  4  atoms  of  TV  form  trititanic  tetra- 

nitride. 

50.  Formation   of  Binaries. — A  binary  molecule  con- 
sists of  atoms  with  either  equal  or  different  valences.     Atoms 
having  the  same  valence  unite  in  the  proportion  of  1    to 
1 ;  that  is,  the  atoms  mutually  saturate  each  other  and  their 
chemism  is  satisfied. 

ILLUSTRATION. — K'  and  C/',  both  monads,  form  K' Cl'  or  K-CL 
S"  and  O",  both  dyads,  form  S"O"  or  S=O. 
Pt™  and  Sziv,  both  tetrads,  form  /VivS/iv  or  Pf^Si. 

When  atoms  with  different  valences  unite,  each  atom  has 
to  furnish  the  same  number  of  bonds  to  satisfy  their  chem- 
ism. This  number  is  the  least  common  multiple  of  the  two 
different  valences.  The  absolute  number  of  atoms  of  each 
constituent  is  obtained  by  dividing"  the  least  common  multi- 
ple by  the  valence  of  each  atom. 

ILLUSTRATION. — When  a  tetrad  and  a  hexad  combine,  each  must 
furnish  12  bonds,  12  being  the  least  common  multiple  of  4  and  6.  To 
do  this,  3  tetrads  and  2  hexads  are  required.  The  following  formulas 
will  further  illustrate  this  law : 

//',  a  monad,  and  O",  a  dyad,  form  H'f>" . 
N1",  a  triad,  and  O",  a  dyad,  form  A™'0'3'. 
Sniv,  a  tetrad,  and  S",  a  dyad,  form  SnivS". 
7V,  a  pentad,  and  O",  a  dyad,  form  /2V<9'5'. 
C/vii,  a  heptad,  and  O",  a  dyad,  form  CY™<9'/. 
6Viv,  a  tetrad,  and  F ,  a  monad,  form  Si® '  F'^. 

Whenever  a  perissad  and  an  artiad  atom  form  a  molecule, 
the  required  number  of  atoms  is  inversely  as  the  valence 
of  each. 

51.  It  was  previously  stated  that  atoms  do  not  exist  in 
a   free   state;    consequently,  the   formation    of    compound 


§  5  THEORETICAL  CHEMISTRY.  39 

molecules  may  be  considered  as  a  process  in  which  an 
exchange  of  atoms  takes  place  between  a  number  of  mole- 
cules that  are  brought  in  close  proximity ;  a  fact  that  may 
easily  be  explained  by  the  theory  that  dissimilar  atoms 
possess  a  far  greater  attraction  for  each  other  than  similar 
atoms. 

ILLUSTRATION.— A  molecule  of  sulphur  S-S  is  brought  under  suitable 
conditions  in  contact  with  a  molecule  of  oxygen  O-O.  The  stronger 
attraction  of  the  atoms  of  oxygen  for  the  atoms  of  sulphur  causes  them 
to  form  a  new  combination,  and  the  result,  in  this  case,  is  SO  +  SO,  or 
2  molecules  of  hyposulphurous  oxide.  Graphically,  it  may  be  repre- 
sented in  the  following  way : 

After  Combination. 

Before  Combination. 


52.     Saturated  .and    Uiisa  titrated    Molecules. — The 

groups  of  atoms  now  being  considered  are  called  saturated 
molecules,  because  the  bonds  of  their  constituent  atoms  are 
all  mutually  engaged.  We  distinguish  between  these  and 
u  nsatnr at ed  groups  of  atoms,  which,  possessing  free  bonds, 
may  enter  into  combination  like  single  atoms.  These  unsat- 
urated  groups  of  atoms  are  known  as  compound  radicals 
(or  radicles).  They  do  not  exist  free  in  nature,  although, 
like  atoms,  -by  combining  with  another  similar  group  they 
may  form  a  saturated  molecule.  Their  valence  is  equal  to 
the  number  of  their  unsatisfied  bonds,  which  is  the  difference 
of  the  valence  of  their  components. 

ILLUSTRATION. — Water  consists  of  H-O-H ';  supposing  1  atom  of 
hydrogen  to  be  removed,  the  unsaturated  group  H-O-  remains,  which, 
though  consisting  of  2  atoms,  is  able  to  enter  into  the  formation  of  a 
molecule  like  any  single  atom.  It  possesses  one  free  bond  and  acts 
therefore  as  a  monad.  It  is  expressed  in  formulaic  form  thus:  (//'(?")'. 
Combining  with  another  similar  group  it  forms  a  saturated  molecule 
(ff'O"}'  +  (N'O")'  =  H-O-O-H,  or  H*O*.  Or  the  pentad  Pv  may  be 
partly  saturated  by  the  dyad  O" ,  forming  the  trivalent  compound 


40  THEORETICAL  CHEMISTRY.  §  5 

radical  (PvO")"f,  which  may  combine  like  a  trivalent  atom,  say  with  67, 
forming  (PvO")'"C/3. 

53.  Names  of  Compound  Radicals. — The  names  of 
compound  radicals  terminate  in  yl.  The  root  of  the  name 
comes  either  from  constituents  of  the  radical  or  from  some 
compound  into  which  it  enters. 

ILLUSTRATION. — The  compound  radical  (H'O")r  is  called  hydroxyl, 
(PVO")'"  is  termQ&phosphoryl,  (Civ7/3)'  is  termed  methyl,  from  methyl 
alcohol,  of  which  it  is  a  constituent.  The  compound  radicals  (H^N}' 
amidogen,  (CN)'  cyanogen,  and  (H^N}'  ammonium  are  the  only 
three  exceptions  to  this  rule. 

Molecules  containing  a  compound  radical,  united  to  an 
atom,  are  regarded  as  binaries  and  named  accordingly. 

ILLUSTRATION. — (NO*)'  and  Cl  form  nitryl  chloride,  (CO)"  and  5  yield 
carbonyl  sulphide,  etc. 


TERNARY  COMPOUNDS. 

54.  The  name  ternary  is  applied   to  those  compounds 
containing  three  elements.     The  third  atom,  which  performs 
a  linking  function,  must  be  a  polyad,  an  atom  whose  valence 
is  greater  than  1,   since  no  monad   can  join,   other  atoms 
together.     Among  the  compounds  higher  than  the  binary 
series  are  those  formed  by  the  union  of  water  with  oxides, 
thus  forming  acids  and  hydrates  (hydroxides)  and  derivatives 
of  these  bodies. 

55.  Acids. — The  name  acid  is  a  familiar  one,  because  it 
is  continually  applied  in  every-day  parlance  to  anything  that 
is  sour.    A  number  of  bodies  possess  this  distinction  in  com- 
mon ;  to  the  chemist  the  sourness  of  an  acid  is  but  an  acci- 
dental  property,   as,    according   to   his   definition    of   these 
bodies,  substances  are  included  as  acids  that  are  not  sour  to 
the  taste.      An  acid  may  be  defined  as  a  body  containing 
hydrogen,  which  hydrogen  may  be  replaced  by  a  metal,   or 
group  of  elements  equivalent  to  a   metal.     It  may  also  be 
defined  as  a  compound  that  will  unite  with  a  base,  forming  a 
salt  and  water.      As  a  class  the  acids  arc  sour;  they  are  also 


§  5  THEORETICAL  CHEMISTRY.  41 

active  chemical  agents.  Most  acids  are  characterized  by  the 
property  of  changing  the  color  of  a  solution  of  litmus,  a 
naturally  blue  body,  to  a  red  tint. 

Oxygen  is  a  constituent  of  most  acids,  the  members  of  this 
group  being  distinguished  as  oxy  acids.  The  oxides  that  by 
union  with  water  form  acids  are  termed  anhydrides,  or 
anhydrous  acids.  They  are  in  most  cases  non-metallic  oxides, 
but  sometimes  consist  of  metals  combined  with  a  compara- 
tively large  number  of  atoms  of  oxygen.  There  are  a  few 
acids  in  which  oxygen  is  absent ;  these  are  termed  hydr acids; 
hydrochloric  acid  HCl  is  an  example  of  this  class.  Accord- 
ing to  the  -definition  given,  hydrogen  is  an  essential  constitu- 
ent of  all  acids.  It  should,  however,  be  mentioned  that 
some  chemists  apply  the  term  "acid"  to  what  are  here 
termed  "anhydrides";  consequently,  CO3  (carbon  dioxide 
or  carbonic  anhydride)  is  sometimes  described  as  carbonic- 
acid  gas.  This  name  is  now,  however,  being  replaced  by 
the  one  more  in  accordance  with  the  theories  of  modern 
chemical  science. 

56.  Bases  and  Alkalies. — The  oxides  that,  when  dis- 
solved in  water,  restore  the  blue  color  to  reddened  litmus 
are  called  alkalies;  they  form  a  subdivision  of  a  larger  class 
of  oxides,  the  whole  of  which  combine  readily  with  acids, 
and  are  known  as  bases. 

A  base  is  a  compound,  usually  an  oxide  or  a  hydrate,  of  a 
metal  (or  group  of  elements  equivalent  to  a  metal),  and  this 
metal  (or  group  of  elements]  is  capable  of  replacing  the  hydro- 
gen of  an  acid,  when  the  two  are  placed  in  contact.  The 
greater  number  of  metallic  oxides  are  bases.  Bases,  as  well 
as  acids,  differ  considerably  in  their  chemical  activity.  As 
already  mentioned,  certain  bases  that  dissolve  in  water  are 
termed  alkalies. 

An  alkali  is  a  base  of  a  specially  active  character.  It  can 
easily  be  recognized  by  its  solubility  in  water,  to  which  it 
imparts  a  soapy  taste  and  feeling,  and  also  by  its  ability  to 
restore  the  blue  color  to  vegetable  blues,  such  as  litmus,  that 
have  been  previously  reddened  by  an  acid.  The  principal 


42  THEORETICAL  CHEMISTRY.  §5 

alkalies  are  sodium  Jiydrate  NaHO  and  potassium  Jiydrate 
KHO.  A  solution  of  ammonia  gas  in  water  is  also  alkaline, 
and  is  often  termed  ammonium  Jiydrate. 

Paper  tinted  yellow  with  turmeric  is  also  used  as  a  test  for 
alkalies,  which  give  the  paper  a  reddish-brown  color. 

The  hydrates  are  mostly  compounds  of  metallic  oxides  with 
water;  they  are  also  sometimes  termed  hydroxides.  (The 
word  hydroxide,  however,  is  objectionable,  because  it 
infringes  on  the  restriction  of  ide  to  the  names  of  binary 
compounds. )  Their  formation  is  represented  by  the  follow- 
ing equations: 

NajO      +       H9O      =      2  NaHO 
sodium  oxide  water  sodium  hydrate 

CaO       4-       HtO       =      Ca(OH\ 
calcium  oxide  water  calcium  hydrate 

57.  Salts. —  When  an  acid  and  a  base  react  on  each  other, 
the  substance  or  compound produ'ced  by  the  replacement  of  the 
hydrogen  of  the  acid  by  the  metal  of  the  base  is  termed  a  salt. 
Water  is  also  formed  during  the  reaction.  The  action  of  the 
stronger  acids  and  bases  on  each  other  is  very  violent;  the 
resulting  salts  are  usually  without  action  on  litmus.  This 
is,  however,  not  always  the  case,  for  when  a  strong  acid 
combines  with  a  weak  base,  the  salt  is  acid  to  litmus,  of 
which  nitrate  of  mercury  is  an  example.  When  the  base  is 
strong  and  the  acid  is  weak,  as  in  sodium  carbonate,  the  salt 
has  an  alkaline  reaction.  The  following  are  instances  of  the 
formation  of  salts  by  the  union  of  acids  and  bases : 

HCl    +     NaHO  =  NaCl  +  H^O 
hydrochloric        sodium          sodium  , 

acid  hydrate        chloride 

HCl     +     NH^HO    =    NHfl  +   HtO 

hydrochloric        ammonium        ammonium 
acid  hydrate  chloride 

T  T  r*  /")         I         /r"Vv  /")  /^"yv  C  /")       I        T  T  S~) 

Jj    o  (_/       — (—      C-«C/     ^^     O#>jC/    — r-    tt    {s 

sulphuric        calcium        calcium 

acid  oxide          sulphate       wai 


§  5  THEORETICAL  CHEMISTRY.  43 

.Litmus,  which  has  been  repeatedly  spoken  of  in  connec- 
tion with  the  ternary  compounds,  is  itself  a  salt  of  a  vegetable 
acid  and  a  vegetable  base  possessing  a  blue  color.  An  acid, 
when  added,  displaces  the  weak  vegetable  acid  and  with  the 
remaining  vegetable  base  forms  a  salt;  the  liberated  litmus 
acid,  being  red,  gives  the  red  tint  to  the  solution.  On  adding 
to  this  solution  another  base,  this  latter  combines  with  the 
stronger  acid,  displacing  in  turn  the  vegetable  base.  The 
displaced  vegetable  acid  and  base  again  unite,  and  the  original 
blue  color  is  also  restored. 

58.  Nomenclature  of  Acids  and  Salts.  —  The  names  of 
acids  are  derived  from  those  of  their  principal  constituents, 
changing  them  into  adjectives  ending  in  ic.  From  sulphur 
we  have  sulpJiuric  acid  ;  from  nitrogen,  nitric  acid  ;  etc. 
Hydr  acids  are  distinguished  by  the  prefix  liydro;  for  instance, 
hydrochloric  acid. 

The  names  of  the  corresponding  salts  are  derived  from  the 
same  root  by  adding  ate.  Thus,  the  salts  of  sulphuric  and 
nitric  acids  are  called  sulphates  and  nitrates,  respectively. 

If  an  element  should  form  two  oxides,  both  of  which 
combine  with  water  to  form  acids,  then  the  acid  contain- 
ing the  higher  proportion  of  oxygen  receives  the  name 
ending  in  ic,  while  for  the  termination  of  the  other  acid  ous 
is  chosen. 

ILLUSTRATION.  —  Two  oxides  of  sulphur  are  known,  S<92,  or  sulphurous 
oxide,  and  SO3,  or  sulphuric  oxide.  Each  combines  with  water  to  form 
acids. 

'  SO,     + 


sulphurous  at  AT         sulphurous 

oxide  acid 

SO3    +     H^O    =   H^SO, 


The  salts  of  acids  whose  names  end  in  ous  have  the  ter- 
mination ite. 

H^SO^    +    ZNaHO  =  Na^SO^  +  2fftO 
sulphurous  sodium  sodium  . 

acid  hydrate      ""  sulphite 


44  THEORETICAL  CHEMISTRY.  §  5 

The  same  termination  that  is  applied  to  the  base  is  also 
applied  to  the  salt.  Thus,  two  different  sulphates  of  mercury 
are  obtained,  namely,  when  sulphuric  acid  is  caused  to  react 
with  mercurous  oxide  and  mercuric  oxide,  respectively. 


H,SOt   +    Hg,0    =    J/g.SOt  +  Hft 

sulphuric        mercurous        mercurous 
acid  oxide  sulphate 

H^SO^  +    HgO    =   HgSO^  +  H,0 

sulphuric        mercuric        mercuric 
acid  oxide  sulphate 

The  salts  of  hydracids,  being  necessarily  binaries,  have 
names  ending  in  ide. 

1HCI     +     HgO    =    HgCl,  +  H^O 

hydrochloric        mercuric        mercuric 
acid  oxide  chloride 


59.  Basicity  of  Acids.  —  In  the  formation  of  salts,  the 
hydrogen  of  an  acid  is  replaced  by  a  metal.  Hydrogen, 
being  univalent,  is  replaced  by  different  metals  in  propor- 
tions varying  according  to  their  valence.  The  valence  of 
the  displacing  metals  must  be  equal  to  that  of  the  total 
number  of  hydrogen  atoms  displaced.  Thus,  a  monad  will 
replace  1  atom  of  hydrogen;  a  dyad,  2  atoms,  and  so  on. 
The  quantity  of  a  base  required  to  form  a  salt  with  an  acid 
depends  on  the  number  of  atoms  of  hydrogen  present  capable 
of  replacement  by  the  base;  the  measure  of  this  quantity  is 
termed  basicity  of  the  acid.  Those  acids  containing  but  1 
atom  of  replaceable  hydrogen  are  termed  monobasic,  those 
with  2  or  3  atoms  of  replaceable  hydrogen  are  called  dibasic 
and  tribasic,  respectively. 

A  monobasic  acid  requires  1  atom  of  a  monad  metal  to 
form  a  salt,  while  with  a  bivalent  metal  2  molecules  of  the 
acid  are  necessary.  In  every  case  such  a  number  of  mole- 
cules of  acid  and  atoms  of  metal  must  be  taken  as  will  give 
the  same  total  valence  of  replaced  hydrogen  atoms  as  that 


§  5 


THEORETICAL  CHEMISTRY. 


45 


of  the  replacing  metal.     The  following  typical  examples  will 
illustrate  this  principle: 


MONOBASIC    ACID. 


HNO*  +  Na'OH  =  Na'  NO*  + 


nitric 
acid 


nitric 
acid 


nitric 
acid 


sodium 
hydrate 


calcium 
hydrate 


bismuth 
hydrate 


sodium 
nitrate 


calcium 
nitrate 


bismuth 
nitrate 


sulphuric 
acid 


sulphuric 
acid 


3//2S<94 

sulphuric 
acid 


DIBASIC   ACID. 

ZNa'OH  =  Na^SO,  + 

sodium  sodium 

hydrate  sulphate 

a"(OH^  =  CaSO<    + 

calcium  calcium 

hydrate  sulphate 


bismuth  bismuth 

hydrate  sulphate 


TRIBASIC   ACID. 


phosphoric 
acid 


phosphoric 
acid 


phosphoric 
acid 


sodium 
hydrate 


sodium 
phosphate 


calcium 
hydrate 


bismuth 
hydrate 


calcium 
phosphate 

=  Bi'"PO, 

bismuth 
phosphate 


6O.  Acid,  Normal,  and  Basic  Salts.  —  Salts  such  as 
those  whose  formulas  are  given  in  the  preceding  examples 
are  known  as  normal  salts.  A  normal  salt  may  be  defined 
as  a  salt  produced  by  the  replacement  of  the  hydrogen  of  the 
acid  by  its  valence  equivalent  of  a  metal,  or  by  a  group  of 
elements  that  acts  as  a  metal,  in  a  base. 

In  the  case  of  those  acids  containing  more  than  1  replace- 
able atom  of  hydrogen,  if  a  portion  only  of  such  hydrogen  is 


46  THEORETICAL  CHEMISTRY.  §5 

replaced,  the  result  is  a  salt  that  still  possesses  an  acid  char- 
acter. An  acid  salt  may,  then,  be  defined  as  a  salt  produced 
by  the  partial  replacement  of  the  hydrogen  atoms  ofapolybasic 
acid  by  the  metal  of  a  base  ;  as,  for  instance: 

H^SO^  +  NaHO  =  NaHSO^  +  HJ) 

sulphuric      sodium        hydrosodium 
acid  hydrate  sulphate 

Certain  oxyacids  exist  that  possess  the  power  of  combining 
with  more  of  the  base  than  is  necessary  to  produce  a  normal 
salt,  thereby  forming  what  are  termed  basic  salts.  A  basic 
salt  may,  then,  be  defined  as  a  salt  produced  by  the  combina- 
tion of  an  acid  with  a  higher  proportion  of  a  base  than  is 
necessary  for  the  formation  of  a  normal  salt.  Yellow  basic 
mercuric  sulphate,  known  as  turpeth  mineral,  is  an  example 
of  a  basic  salt,  the  additional  quantity  of  base  being  distin- 
guished here  by  being  placed  after  the  comma  in  the  formula 
Its  formula  is  generally  written  thus: 


61.  Classification  of  Acids.  —  Oxyacids  are  sometimes 
divided  into  ortho-  and  meta-acids,  but  this  distinction  is  not 
important  and  these  terms  are  often  used  in  a  rather  loose 
way.  In  its  strictest  sense,  an  ortho-acid  is  one  containing 
as  many  hydroxyl  groups  as  the  maximum  valence  of  the 
principal  element.  Thus,  phosphorus,  having  a  maxi- 
mum valence  of  5  (as  is  shown  by  the  formation  of  PCl^), 
might  be  expected  to  form  an  acid  having  the  formula 
P(OH)b,  and  this  would  be  called  ortho-phosphoric  acid.  As 
a  matter  of  fact,  this  acid  is  not  known,  and  only  a  very  few 
acids  formed  in  this  way  do  exist  ;  hence,  the  term  is  often 
used  in  a  broader  sense  and  is  applied  to  the  normal  tri- 
basic  acids.  Thus,  PO(OH)Z  is  frequently  called  ortho- 
phosphoric  acid. 

A  meta-acid  is  one  formed  by  extracting  water  from  an 
ortho-acid.  Hence,  if  we  use  the  term  ortho  in  its  strictest 
sense  in  the  case  just  mentioned,  and  call  P(OH\  ortho- 
phosphoric  acid,  the  acid  usually  called  ortho-  phosphoric 
PO(OH)^,  which  would  be  formed  by  removing  one  molecule 


§  5  THEORETICAL  CHEMISTRY.  47 

of  water  from  P(OH}^  would  be  a  meta-acid;  and  by  remov- 
ing another  molecule  of  water  we  would  have  another 
meta-acid  PO^OH),  which  is  usually  known  as  meta-plios- 
phoric  acid.  Thus,  we  would  have  two  meta-acids  of  phos- 
phorus. To  avoid  confusion,  when  the  term  ortJio  is  used 
in  its  strictest  sense,  the  term  anhydro-acids  is  sometimes 
applied  to  those  obtained  by  the  removal  of  water  from  the 
ortho-acids,  and  the  term  meta  is  reserved  for  a  particular 
acid.  This  distinction  is  only  important  in  studying1  the 
theoretical  formation  of  acids,  but  the  terms  ortho  and  meta 
are  important  in  organic  chemistry  where  they  are  applied 
in  a  different  manner. 


VOLUME  RELATIONS  OF  MOLECULES. 

62.  Molecular    Volume. — The    fact    that    all    gases, 
whether  elementary  or  compound,  expand  and  contract  at 
the  same  rate  when  subjected  to  variations  of  temperature 
and  pressure,  has  an  important  bearing  on  their  probable 
molecular  conditions.     Their  similarity  in  this  respect  has 
led  to  the  assumption  expressed  in  Avogadro's  law  (see  Art. 
24),  from  which  it  follows  that,  at  the  same  temperature 
and  under  the  same  pressure,  the  volume  of  any  gaseous 
molecule  is  the  same,  whatever  may  be  the  nature  and  com- 
position of  the  gas.     Every  molecular  formula,   therefore, 
not  only  expresses  the  weight  but  also  the  volume  of  the 
molecule.      The    molecule   of    hydrogen    is   chosen   as   the 
standard  of  molecular  volume ;  but  as  it  is  sometimes  con- 
venient  to   speak   of   atomic  volume — which,    however,    is 
inconsistent   since   atoms   do    not   exist    free — the   volume 
occupied  by  an  atom  of  hydrogen  is  taken  as  unity ;  the  vol- 
ume occupied  by  a  molecule   of  hydrogen  will  be  2,   and, 
since  all  molecules  occupy  the  same  volume,  the  volume  of  a 
molecule  of  any  gaseous  body  is  assumed  to  be  2,  also,  when 
this  system  is  used. 

63.  Density  and  Molecular  Weight. — The  density  of 
a  gas  has  already  been  defined  as  the  weight  of  any  volume 


48  THEORETICAL  CHEMISTRY.  §  5 

compared  with  that  of  the  same  volume  of  hydrogen  (see 
Art.  26).  As  the  molecule  of  hydrogen  contains  two  atoms, 
its  molecular  weight,  expressed  in  terms  of  atomic  weight, 
is,  consequently,  2.  The  molecular  weight  of  any  body  in 
the  gaseous  state  is  the  weight  of  that  volume  which  occu- 
pies the  same  space  as  do  2  parts  by  weight  of  hydrogen. 
Conversely,  as  the  molecular  weight  is  the  sum  of  the 
weights  of  the  constituent  atoms,  the  density  of  a  gas  may 
be  determined  from  its  formula.  As  the  molecule  of  hydro- 
gen weighs  2,  the  density  of  any  gas,  whether  elementary  or 
compound,  is  identical  with  the  half  of  its  molecular  weight. 
For  instance,  the  molecule  of  steam  weighs  1 8,  and  occupies 
the  same  space  as  the  molecule  of  hydrogen,  which  weighs 
2 ;  the  quantity  of  steam  occupying  the  same  space  as  1  vol- 
ume -of  hydrogen  must,  therefore,  weigh  ±£-  =  9. 

64.  Absolute  Weight  of  Hydrogen. — Hydrogen  being 
taken  as  the  unit  of  density,  its  absolute  weight  is  of  great 
importance.  As  a  result  of  most  careful  weighing,  it  has 
been  found  that  1  liter  of  hydrogen  at  the  normal  tempera- 
ture and  pressure  weighs,  approximately,  .0896  gram,  or 
11.16  liters  weigh  1  gram.  It  is  absolutely  necessary  for 
the  student  to  remember  that  the  weight  of  1  liter  of  hydro- 
gen at  0°  and  760  millimeters  pressure  is  .0896  gram,  and  he 
should  therefore  thoroughly  impress  it  on  his  mind.  Know- 
ing this,  there  is  absolutely  no  difficulty  in  calculating  the 
weight  of  any  other  gas  whose  composition  is  known.  The 
weight  in  grams  of  a  liter  of  any  gas  is  its  density  multiplied 
by  .0896;  further,  the  weight  in  grams  of 11.16  liters  of  any 
gas  is  identical  with  the  number  representing  its  density. 
The  density  of  oxygen,  for  instance,  being  16,  the  weight  of 
1  liter  at  N.  T.  P.*  is  .0896x16  =  1.4336  grams.  The 
weight  of  a  compound  gas  is  found  just  as  easily.  The  den- 
sity of  hydrochloric  acid  is  18.18;  therefore,  the  weight  of  1 
liter  of  this  acid  is  .0896  X  18.18  —  1.6289  grams,  or  11.16 
liters  weigh  18.18  grams. 


*  N.  T.  P.  is  used  as  abbreviation  of  normal  temperature  and  pres- 
sure. 


THEORETICAL  CHEMISTRY. 


40 


This  constant,  .0896,  is  of  such  frequent  occurrence  in 
chemical  calculations  that  it  has  been  proposed  to  give  it  a 
distinct  -name,  crith  ;  the  weight  of  a  liter  of  any  other  gas 
is  then  expressed  as  so  many  criths. 

65.  Diffusion  of  Gases. — Diffusion  is  the  intermixture 
of  molecules  brought  about  by  their  power  of  moving  against 
one  another.  This  power  is  possessed  in  the  highest  degree 
by  gaseous  molecules,  all  gases  being  capable  of  perfect  and 
comparatively  rapid  intermixture.  Some  liquids,  such  as 
alcohol  and  water,  also  intermix  perfectly,  although  com- 
paratively slowly ;  while  other  liquids,  such  as  oil  and  water, 
diffuse  into  each  other  only  to  a  very  limited  extent.  When 
alcohol  is  poured  on  to  water  it  forms  a  separate  layer  on  the 
surface  of  the  water,  because  it  is  specifically  the  lighter  of 
the  two.  'After  a  time,  however,  the  two  layers  will  no 
longer  be  discernible,  and  the  liquid  will 
be  a  homogeneous  mixture  of  alcohol  and 
water.  In  the  same  way,  hydrogen  will 
float  on  air,  but  only  for  a  very  short  time, 
since  the  rate  of  diffusion  of  this  gas  is  ex- 
tremely rapid.  A  homogeneous  mixture  of 
hydrogen  and  air  will  be  speedily  formed. 

The  molecules  of  all  gases  are  constantly 
moving,  but  they  do  not  move  with  the 
same  velocity,  so  that  some  gases  diffuse 
more  rapidly  than  others.  This  has  been 
discovered  by  confining  a  gas  in  a  vessel 
closed  by  some  material  having  very  .minute 
pores,  and  immersing  the  vessel  in  an 
atmosphere  of  some  other  gas.  The  pas- 
sage of  the  molecules  through  the  pores  of 
the  material  closing  the  vessel  is  sufficiently 
slow  to  allow  of  a  comparison  between  the 
velocities  of  passage  or  rates  of  diffusion  of  the  two  gases. 

For  this  purpose,  a  diffusion  tube  (see  Fig.  5)  is  employed. 
It  consists  of  a  glass  tube  A,  closed  at  one  end  by  a  plate  of 
piaster  of  Paris  B.  If  this  tube  is  filled  with  hydrogen  and 


FIG.  5. 


50  THEORETICAL  CHEMISTRY.  §  5 

its  open  end  immersed  in  colored  water,  the  water  will  be 
observed  to  rise  rapidly  in  the  tube  on  account  of  the  rapid 
escape  of  the  hydrogen  through  the  pores  of  the  plaster  of 
Paris.  The  external  air,  of  course,  passes  into  the  tube 
through  the  pores  at  the  same  time,  but  much  less  rapidly 
than  the  hydrogen  passes  out,  so  that  the  ascent  of  the  column 
of  water  C  marks  the  difference  between  the  volume  of 
hydrogen  that  passes  out  and  the  volume  of  air.  that  passes 
into  the  tube  in  a  given  time,  and  allows  a  measurement 
to  be  made  of  the  rate  of  diffusion,  that  is,  of  the  velocity 
with  which  the  gas  issues  as  compared  with  the  velocity 
with  which  the  air  enters,  this  velocity  being  always  taken 
as  unity.  To  determine  the  rate  of  diffusion  it  is,  of  course, 
necessary  to  maintain  the  water  at  the  same  level  within 
and  without  the  diffusion  tube,  so  as  to  exclude  the 
influence  of  pressure. 

This  method  has  disclosed  the  law  of  diffusion,  known  as 
Graham's  laiv,  namely,  that  the  rate  of  diffusion  of  gases  is 
inversely  as  the  square  root  of  tlieir  density.  Thus,  hydro- 
gen and  oxygen  have,  respectively,  densities  of  1  and  10; 
therefore,  hydrogen  diffuses  four  times  as  rapidly  as  oxygen. 

66.  Combination  by  Volume. — Gay-Lussac,  who 
investigated  the  proportions  in  which  gaseous  volumes 
enter  into  combination,  established  the  following  laws: 

1.  The  ratio  in  which  gases  combine  by  volume  is  always 
a  simple  one. 

2.  The  volume  of  the  resulting  gaseous  product  bears  a 
simple  ratio  to  the  volume  of  its  constituents. 

The  laws  of  combination  by  volume,  which  at  the  time  of 
Gay-Lussac's  investigation  (between  1805  and  1808)  were 
simply  experimental,  have  recently  been  investigated  by 
Clausius,  who  has  shown  that  they  are  really  nothing  but  a 
very  simple  deduction  from  the  law  of  Avogadro. 

According  to  Avogadro's  law,  equal  volumes  of  all  gaseous 
bodies,  contain  the  same  number  of  molecules.  If,  there- 
fore, the  number  of  molecules  is  in  some  way  decreased,  the 


§  5  THEORETICAL  CHEMISTRY.  51 

volume  must  necessarily  be  decreased  also.  If  we  assume 
that  in  a  given  volume  of  any  gas  each  molecule  is  diatomic, 
and  that  by  some  means  the  molecule  can  be  made  tetratomic 
—the  number  of  atoms  remaining  the  same — the  number  of 
molecules  will  be  reduced  one-half,  since  each  molecule  con- 
tains twice  the  number  of  atoms;  and  this  reduction  of 
molecules  is  naturally  accompanied  by  a  corresponding 
decrease  of  the  volume ;  that  is,  the  volume  of  the  gas  will 
also  be  reduced  one-half. 

Or,  if  we  assume  that  the  diatomic  molecule  could  be 
made  triatomic,  the  number  of  molecules,  and,  consequently, 
the  volume,  would  be  reduced  one-third. 

To  apply  this  argument  to  the  facts  of  volume  combination, 
we  will  consider  separately  the  combination  of  hydrogen  with 
monads,  dyads,  triads,  and  tetrads,  assuming  all  their  mole- 
cules to  be  diatomic. 

1.  Combination  of  a  Monad  ivitli  Hydrogen. — Each 
atom  of  a  monad  combines  with  an  atom  of  hydrogen; 
before  they  combine,  their  molecules  are  each  .diatomic, 
and  each  molecule  of  the  compound  is  also  diatomic;  in 
accordance  with  the  principles  of  Art.  24,  if  there  was 
1  volume  of  each  before  the  combination,  the  number  of 
compound  molecules  after  the  combination  takes  place  is 
exactly  equal  to  the  sum  of  the  molecules  of  the  monad 
and  hydrogen  before  they  combined,  and  the  result  is  2 
volumes  of  the  compound. 

Monads,  then,  combine  with  one  another,  volume  to  vol- 
ume, yielding  2  voluirfes  of  the  product. 

ILLUSTRATION. — The  monad  atom  chlorine  Cl  and  the  monad  atom 
hydrogen  //"combine  to  form  hydrogen  chloride  HCL  Their  molecules 
being  diatomic,  they  unite,  molecule  to  molecule,  volume  to  volume, 
and,  consequently,  the  number  of  molecules  and  the  volume  they 
occupy  will  remain  exactly  the  same  after  combination  has  taken  place 
as  it  was  prevfous  to  it. 

We  may  represent  this  molecularly  thus : 

H  Cl  H-  Cl 

|       +      |      yield 
H  Cl  H-Cl, 


52  THEORETICAL  CHEMISTRY. 

and  volumetrically  thus: 


and 

C19 

form 

(HCl)9 

2.  Combination  of  a  Dyad  with  Hydrogen. — A  dyad  atom 
will  require  2  atoms  of  hydrogen ;  1  molecule  will  require  2 
molecules  of  hydrogen ;  and  1  volume,  2  volumes  of  hydro- 
gen.    Or,  to  express  this  fact  in  other  words,  dyads  combine 
with  monads  in  the  ratio  of  1  to  2. 

As  the  resulting  molecule  is  triatomic,  it  is  evident  that  3 
diatomic  molecules  have  combined  to  form  2  triatomic  mole- 
cules, and  that  the  decrease  in  the  number  of  molecules  also 
causes  a  corresponding  decrease  in  volume.  Thus  we  see 
that  3  volumes  of  simple  gas  give  2  volumes  of  compound 
gas,  a  condensation  of  3  volumes  to  2  volumes  taking  place 
during  combination. 

Dyads,  then,  combine  with  monads  in  the  ratio  of  1  to  2, 
and  yield  2  volumes  of  the  product. 

ILLUSTRATION. — An  atom  of  oxygen  is  bivalent,  and  its  molecule  dia- 
tomic. One  atom  of  oxygen  requires  2  atoms  of  hydrogen  to  form  a 
molecule  of  water  gas,  and  1  molecule  of  oxygen  requires  2  molecules 
of  hydrogen,  and  1  volume  requires  2  volumes.  This  may  be  repre- 
sented molecularly  thus : 

Before  combination.  After  combination. 
HOH  H-O-H 

I   II   I  H-O-H 

HOH 

3.  Combination  of  a   Triad  with* Hydrogen. — One  triad 
atom  requires  3  monad  atoms  for  saturation,    1   molecule 
requires  3  molecules,  and  1  volume  requires  3  volumes;  since 
the  newly  produced  molecule  is  tetratomic,  the  number  of 
molecules  and  the  corresponding  volume  they  occupy  must 
be  reduced  one-half. 

Triads,  then,  unite  with  monads  in  the  ratio  of  1  to  3, 
yielding  2  volumes  of  the  product. 

ILLUSTRATION. — The  trivalent  atom  of  nitrogen  TV  requires  3  atoms 
of  hydrogen  to  form  1  molecule  of  ammonia. 


THEORETICAL  CHEMISTRY 

N  HHH  HHH 

III       +         III  \1X       + 

N  HHH  N 


53 


or 


JB, 


+    B, 


As  4  diatomic  molecules  yield  2  tetratomic  molecules,  1 
volume  of  nitrogen  and  3  volumes  of  hydrogen  will  yield 
2  volumes  of  ammonia  gas. 

4.  Combination  of  a  Tetrad  with  Hydrogen.  —  One  tetrad 
atom  requires  4  monad  atoms  for  saturation  ;  1  tetrad  mole- 
cule will  require  4  monad  molecules,  and  1  volume  of  any 
tetrad  will  require  4  volumes  of  any  monad;  as  the  resulting 
molecule  consists  of  5  atoms,  the  5  original  volumes  will  be 
condensed  to  two. 

Tetrads,  then,  unite  with  monads  in  the  ratio  of  1  to  4  and 
yield  2  volumes  of  the  product. 

ILLUSTRATION.  —  The  tetrad  atom  of  carbon  C  combines  with  4  monad 
atoms  of  hydrogen  to  form  marsh  gas.  That  is,  5  diatomic  molecules 
unite  to  form  2  pentatomic  molecules,  or,  as  in  this  case,  1  volume  of 
carbon  and  4  volumes  of  hydrogen  yield  2  volumes  of  marsh  gas. 


HHHH 

|    |    |    | 

HHHH 


H 

| 
C 

\ 
H 


H 

| 
H-C-H 

| 
H 


The  preceding  combinations  may  be  graphically  repre- 
sented thus: 


(2) 


THEORETICAL  CHEMISTRY. 


(3) 


(1)  1  monad  molecule  and  1  monad  molecule  yield  2  diatomic  mole- 
cules. 

(2)  1  dyad  molecule  and  2  monad  molecules  yield  2  triatomic  mole- 
cules. 

(3)  1  triad  molecule  and  8  monad   molecules   yield   2  tetratomic 
molecules. 

(4)  1  tetrad  molecule  and  4  monad  molecules  yield  2  pentatomic 
molecules. 


67.  There  are  two  more  cases  to  be  considered,  namely, 
where  the  molecules  are  either  monatomic  or  tetratomic. 

1.  As  all  known  monatomic  molecules  are  dyads,  the 
combination  with  monad  atoms  would  be : 

1  atom  and  2  atoms  give  3  atoms ; 

1  molecule  and  1  molecule  give  1  molecule; 

1  volume  and  1  volume  give  1  volume ; 

or,  in  other  words,  monatomic  dyad  molecules  combine  with 
diatomic  monad  molecules  in  equal  volume  and  yield  1  vol- 
ume of  the  product. 

ILLUSTRATION. — The  monatomic  dyad  zinc  Zn  combines  with  the 
diatomic  monad  chlorine  Cl ;  that  is,  1  atom  of  zinc  combines  with 


THEORETICAL  CHEMISTRY. 


55 


2  atoms  of  chlorine,  or  1  molecule  of  zinc  combines  with  1  molecule  of 
chlorine,  or  1  volume  of  zinc  with  1  volume  of  chlorine.  The  product 
is  the  triatomic  molecule  zinc  chloride  Z.nCli,  a  condensation  of  half  the 
original  volume  taking  place. 

2.  As  all  known  tetratomic  molecules  are  triads,  the 
atomic  combination  is: 

1  atom  and  3  atoms  yield  4  atoms ; 
1  tetratomic  molecule  and  6  diatomic  molecules  yield 

4  tetratomic  molecules ; 

1  volume  and  6  volumes  yield  4  volumes; 

which  means  that  tetratomic  triad  molecules  combine  with 

diatomic  monad  molecules  in  the  ratio  of  1  to  6  volumes, 

producing  4  volumes  of  the  product. 

ILLUSTRATION. — Phosphorus  is  a  triad,  possessing  a  tetratomic  mole- 
cule. When  it  unites  with  chlorine,  we  have,  atomically,  P  and  C/3 
giving  PC/a,  and  molecularly,  /*4  and  6C/2  giving  (/*C/3)4,  or  by  vol- 


cia 

C12 

Cl, 

Clg 

Cl. 

C19 

or,  as  it  is  also  sometimes  graphically  expressed, 


Cl, 

Cl, 

CHEMICAL  CALCULATIONS. 


REACTIONS. 

68.  Molecular  Stability. — All  molecules  are  more  or 
less  liable  to  chemical  change.  Their  atoms  may  be  altered 
in  number,  in  kind,  or  in  relative  position  by  various  exter- 
nal influences.  A  molecule  is  the  more  stable  in  proportion 
as  it  resists  this  tendency  to  change. 


5G  THEORETICAL  CHEMISTRY.  §5 

69.  Chemical  Keactioiis  and   Reagents.  —  The    term 
chemical  reaction  is  applied  to  any  change  that  takes  place 
in  the  atoms  composing  a  molecule.     The  .substance  applied 
to  produce  the  change  is  called  a  reagent. 

70.  Chemical  reaction  always  occurs  within  the  mole- 
cule.    Hence,  when  two  substances  react  on  each  other,  the 
changes  that  result  maybe  considered  as  taking  place  between 
single  molecules  ;  and  as  all  molecules  in  homogeneous  matter 
are  alike,  and  what  is  true  of  one  molecule  must  be  true  for 
any  mass  of  them,  it  follows  that  a  molecular  change  repre- 
sents a  mass  change,  also. 

7  1  .  Chemical  Equations.  —  Since  every  chemical 
change  or  reaction  is  simply  an  alteration  in  the  association 
and  position  of  atoms  within  the  molecule^  and,  consequently, 
a  change  of  the  constitution  of  the  molecule,  this  reaction 
may  be  expressed  by  an  equation. 

The  substances  entering  into  the  reaction  are  called  the 
factors;  those  issuing  from  the  reaction  are  called  the 
products. 

The  equation  representing  the  reaction  is  written  accord- 
ing to  the  following  rule  : 

Place  the  formulas  of  the  factors  —  connected  by  the  sign 
of  plus  —  as  t  lie  first  member  of  the  equation,  and  the  formulas 
of  the  products  —  also  connected  by  the  sign  of  plus  —  as  the 
second. 

ILLUSTRATION.  —  The  reaction  of  the  2  molecules  A  B  and  CD  would 
then  be  expressed  as  follows  : 

A  B  +  CD  =  A  D  +  B  C. 


72.  A  chemical  equation,  however,  not  only  expresses 
the  fact  of  reaction,  but  it  also  indicates  the  quantities,  by 
weight  concerned  in  it,  and  since  the  atoms  remain  exactly 
the  same  after  the  reaction  as  they  were  before,  being  only 
differently  associated,  it  is  evident  that  a  loss  of  weight 
cannot  occur  through  chemical  reaction,  and  the  weight  of 
the  bodies  resulting  from  a  chemical  change  must  be  the 


§  5  THEORETICAL  CHEMISTRY.  57 

same  as  that  of  the  bodies  before  the  change,  whatever  it 
may  be,  had  occurred. 

73.  Chemists  classify  chemical  reactions  in  the  following 
manner: 

Analytical  reactions  represent  the  separation  of  a  complex 
molecule  into  more  simple  ones. 

ILLUSTRATION.—          2//2<9    =    2//2      +      O2 

water          hydrogen        oxygen 

Synthetical  reactions  represent  the  union  of  simple  mole- 
cules to  form  one  or  more  complex  ones. 

ILLUSTRATION.—      H*      +      C/2      =      1HCI 

hydrogen        chlorine         hydrochloric 

Metathetical  reactions  represent  an  exchange  or  transpo- 
sition of  atoms  between  molecules. 

ILLUSTRATION.—    Pb(NO-^   +   Na^SC*    =    Pb(SO*)   +   2NaNO3 

lead  sodium  lead  sodium 

nitrate  sulphate  sulphate  nitrate 

74.  Facility  of  Chemical  Reaction. — Since  a  chemical 
reaction  is  the  result  of  the  reciprocal  action  of  atoms  and  has 
for  its  effect  a  change  in  the  composition  of  the  molecule,  it  is 
evident  that  it  can  only  take  place  when  these  atoms,  and, 
consequently,  their  molecules,  are  brought  into  intimate  rela- 
tion, or,  more  exactly,  when  the  molecules  of  one  of  the  bodies 
enter  within  the  sphere  of  action  of  other  bodies.     This  sphere 
is  rather  limited,  as  the  affinity  of  the  atoms  is  only  exercised 
at  infinitely  small  distances.     In  consequence,  affinity  is  often 
retarded  by  cohesion,  which  maintains  the  relations  between 
molecules  of  solid  bodies.     These  forces  are  frequently  in 
opposition,  and  in  order  that  the  first  may  obtain  supremacy 
it  is  necessary  that  the  other  shall  yield.     To  increase  the 
affinity  between  two  or  more  bodies  it  becomes  necessary  to 
diminish   their  cohesion.     On  this  condition  the  molecules 
can  enter  within  the  sphere  of  their  reciprocal  attraction,  and 
the  atoms  of  one  body  can  attract  those  of  the  other. 


58  THEORETICAL  CHEMISTRY.  §  5 

Experience  teaches  us  that  this  cohesion  can  to  a  certain 
extent  be  overcome  and  chemical  reaction  facilitated  when 
the  acting  substances  are  brought  into  the  liquid  or  gaseous 
state.  Hence,  either  heat  by  which  bodies  are  vaporized,  or 
solution,  by  which  bodies  are  liqtiened,  is  an  important  aid 
in  producing  chemical  reaction. 

75.  Precipitation. — If  we  place  some  solution  of  mer- 
curic chloride  in  a  test  tube  and  add  to  it,  drop  by  drop,  an 
iodide  of  potassium  solution,  a  red  powder  is  formed,  which 
gradually  settles  down  to  the  bottom  of  the  tube.     This  pow- 
der is  mercuric  iodide,  while  potassium  chloride  remains  in 
solution. 

Whenever  a  substance  separates  from  a  solution  through 
the  addition  of  another,  as  the  mercuric  iodide  has  done  in 
the  example  cited,  the  term  precipitate  is  applied  to  the  sep- 
arated substance,  and  the  substance  that  will  produce  a  pre- 
cipitate is  known  as  a  precipitant.  The  process  of  producing 
a  precipitate  is  known  as  precipitation. 

Those  conditions  of  chemical  change  depending  upon  solu- 
bility are  stated  in  Berthollet's  /aw,  which  may  be  expressed 
as  follows  : 

When,  on  mixing  two  substances  in  solution,  a  new  com- 
pound can  be  formed  that  is  insoluble  in  tJie  solvent  employed, 
such  compound  will  be  formed  and  will  appear  as  a 
precipitate. 

76.  Another  law,  also  established  by  Berthollet,  relates 
to  the  products  of  reactions  that  are  volatile  instead  of  being 
insoluble  solids ;  it  may  be  stated  as  follows : 

When,  on  mixing  different  substances,  a  new  substance  that 
is  volatile  can  be  produced  by  the  rearrangement  of  the  atoms 
of  the  partaking-  substances,  such  new  substance  ivill  be  pro- 
duced and  will  appear  as  a  gas. 

ILLUSTRATION. — If  we  mix  sodium  nitrate  NaNO3  and  sulphuric  acid 
fJ-2SO*  together,  through  a  rearrangement  of  the  atoms  of  these  two 


§  5  THEORETICAL  CHEMISTRY.  59 

substances,  hydrosodium  sulphate  and  nitric  acid  are  produced,  accord- 
ing to  the  equation  : 


NaNO3  +  HtSOt  =  HNO3  +  HNaSO* 

sodium          sulphuric         nitric         hydrosodium 
nitrate  acid  acid  sulphate 

Nitric  acid,  being  volatile,  escapes  as  a  gas  when  heat  is  applied. 

By  means  of  Table  8,  which  shows  the  solubility  of 
various  compounds,  the  resulting  chemical  change  of  these 
compounds,  covered  by  the  first-mentioned  law  of  Berthollet, 
may  be  predicted,  and  as  the  student  progresses  in  his  study 
and  experimenting  he  will  soon  acquire  sufficient  knowledge 
to  predict  also  the  changes  covered  by  the  second-mentioned 
law  of  Berthollet,  which  requires  some  familiarity  with  the 
volatility  of  the  various  compounds. 

ILLUSTRATION.  —  The  reactions  obtained  by  mixing  calcium  chloride 
and  sodium  carbonate  may  be  expressed  by  the  following  equation  : 

Cad*  +  Na^COs  =  CaCO3  +  ZNaCl 

calcium         sodium  calcium          sodium 

chloride       carbonate        carbonate       chloride 

Will  there  be  a  precipitate  ? 

Consulting  Table  8,  we  find  that  calcium  carbonate  is  insoluble, 
and  we  can  predict  that  under  the  given  conditions  calcium  carbonate 
will  be  obtained  in  the  solid  form  as  a  precipitate. 

The  reactions  obtained  by  mixing  calcium  hydrate  and  carbon 
dioxide  may  be  expressed  by  the  equation  : 

Ca(OH}*  +  CO*    =    CaCO3  +  H*O 

calcium         carbonic         calcium 
hydrate          dioxide        carbonate 

Will  there  be  a  precipitate  ? 

Consulting  the  table,  we  find  that  calcium  carbonate  is  insoluble  in 
water,  and,  therefore,  we  can  predict  that  under  the  given  conditions 
calcium  carbonate  will  be  obtained  as  a  precipitate. 


77.  Modes  of  Chemical  Action. — Although  chemical 
actions  vary  considerably  in  their  character,  they  can  all  be 
classified  under  the  following  five  heads : 

1.     Direct  Union  or  Synthesis. — 
EXAMPLE.—  Zn    +     C/2     =     ZnCl* 

_:„„ 

zinc        chlorine 


fiO 


THEORETICAL  CHEMISTRY. 


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•ouny 


v  is  •£  o  d 
W  S  6  <l  O.o'o 


.          rt 

n  I 

O  r3 


THEORETICAL  CHEMISTRY 


61 


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VJ  § 

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^  o 


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enit 
oma 


O.T:     VH  .C 

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62  THEORETICAL  CHEMISTRY.  §  5 

2.  Chemical  Displacement.  —  The  various  elements  possess 
different   degrees   of   chemical   activity;    thus,    magnesium 
enters  into  combination  with  other  elements  with  great  read- 
iness, while  platinum,  on  the  other  hand,  is  almost  inactive, 
forming  but  few  compounds.      In  consequence  of  such  differ- 
ences in  activity,  some  elements  are  able  to  displace  others 
from   compounds  already  formed.     Thus,   iron   is   a   more 
active  element  than  copper,  and  if  a  piece  of  it  is  placed  in  a 
solution  of  a  compound  of  copper  and  chlorine,  the  iron  dis- 
places  the   copper;    a   compound   of   iron   and   chlorine   is 
formed,  and  copper  is  liberated  in  the  free  state.     Again, 
chlorine  is  a  more  active  element  than  oxygen,  and  if  water 
—  a  compound  of  oxygen  and  hydrogen  —  is  submitted  to  the 
action  of  chlorine,  the  latter  element  displaces  the   oxygen, 
and  a  compound  of  hydrogen  and  chlorine  is  the  restilt. 

3.  Mutual  Exchange.  —  When  two  or  more  compounds  are 
brought  together,  there  are  sometimes  conditions  that  lead 
to  an  exchange  between  them  of  their  various  constituents. 
Thus,  if  a  solution  of  mercuric  chloride  (a  compound  of  mer- 
cury and  chlorine)  and  potassium  iodide  (a  compound   of 
potassium  and  iodine)  are  mixed  together,  the   mercury  and 
iodine,  through  mutual  attraction,  unite  to  form  a  new  com- 
pound —  mercuric  iodide.      So,  too,  the  remaining  elements, 
potassium  and  chlorine,  also  unite  to  form  a  new  compound 
—potassium  chloride.     Putting  it  in  other  words,  the  mer- 
cury has,  under  these  conditions,  a  greater  attraction  for 
iodine   than   for   chlorine,  and,    consequently,   the   mercury 
takes  the  iodine  and  hands  over  the  chlorine  to  the  potassium. 

The  chemical  change  may  be  represented  thus  : 

mercuric  j  mercury  —  -  mercury  [  mercuric 
chloride    ]  chlorine  "'^v.  ^^    iodine    j     iodide 

potassium  f  potassium          ^^"^          potassium  )  potassium 
iodide  iodine     ^^  ^^^  chlorine        chloride 


4.  Rearrangement  of  Atoms.  —  This  class  of  chemical 
action  belongs  almost  exclusively  to  the  domain  of  organic 
chemistry.  There  are  many  examples  in  which,  without 
altering  the  proportions  of  each  element  present,  fresh  com- 
pounds may  be  formed  by  changing  the  arrangement  of  the 


§  5  THEORETICAL  CHEMISTRY.  63 

atoms;  that  is,  their  relative  position  within  the  molecule. 
Such  changes  may  be  compared  to  those  in  which  the  same 
letters  by  transposition  may  be  caused  to  form  two  different 
words. 

5.  Direct  Decomposition.  —  The  separation  of  complex 
molecules  into  simpler  ones,  to  which  all  analytical  reactions 
belong;  as,  for  instance: 

=  ff0        S0 


sulphuric  sulphuric 

acid  wate]         oxide 


STOICHIOMETKY. 

78.  Stoichiometry  considers  the  numerical  relations  of 
atoms,  and  all  calculations  that  can  be  based  on  the  atomic 
weights  and  volumes  are  stoichiometrical  calculations.    They 
are,  consequently,  of  the  greatest  importance  in  analytical 
and  synthetical  work. 

79.  We  have  seen  previously  that  each  atom  has  its  own 
weight,    called   the   atomic  weight,    and  that   these    atomic 
weights  are  the  smallest  portions  by  weight  of  any  simple 
substance — referred  to  the  atom  of  hydrogen — that  can  take 
part  in  a  chemical  change. 

A  molecule  being  formed  of  atoms,  a  molecular  weight  is 
the  sum  of  the  weights  of  the  atoms  of  which  it  is  composed. 
It  is  also  equal  to  twice  the  weight  of  a  given  volume  of  a 
substance  in  the  state  of  vapor,  compared  with  the  same 
volume  of  hydrogen  gas. 

80.  Every  chemical  reaction  is  simply  an  alteration  in 
the  association  and  position  of  atoms,  and  all  chemical  equa- 
tions that  represent  such  a  change  represent  it  as  occurring 
between  definite  quantities  of  matter. 

As  an  example,  the  following  equation  represents  the 
chemical  action  occurring  when  solutions  of  mercuric  chloride 
and  potassium  iodide  are  mixed  together. 

HgCl,    +    2AT7    =     Hgl^    +    ^KCl 

mercuric       potassium       mercuric       potassium 
chloride  iodide  iodide  chloride 


64  THEORETICAL  CHEMISTRY.  §  5 

With  the  assistance  of  a  table  of  atomic  weights,  the 
chemist  learns  from  this  equation  that  1  part  of  mercuric 
chloride,  containing  1  combining  proportion  of  mercury, 
weighing  200,  and  2  combining  proportions  of  chlorine,  each 
weighing  35.5,  together  with  2  parts  of  potassium  iodide, 
each  containing  1  combining  proportion  of  potassium  weigh- 
ing 39,  and  1  of  iodine  weighing  127,  together  yield,  or  pro- 
duce, 1  part  of  mercuric  iodide,  containing  1  combining 
proportion  of  mercury,  weighing  200,  and  2  of  iodine,  weigh- 
ing 127  each,  and  2  parts  of  potassium  chloride,  each  con- 
sisting of  1  combining  proportion  of  potassium,  weighing  39, 
and  1  of  chlorine  weighing  35.5.  As  no  chemical  change 
affects  the  weight  of  matter,  the  weight  of  the  quantity  of  a 
compound  represented  by  its  formula  must  be  the  sum  of 
those  of  the  constituent  atoms;  so,  too,  the  weight  of  the 
bodies  resulting  from  a  chemical  change  must  be  the  same 
as  that  of  the  bodies  before  the  change,  whatever  it  may  be, 
had  occurred.  Also  the  same  number  of  combining  propor- 
tions of  each  element  must  appear  on  each  side  of  the  equa- 
tion. Although,  given  a  chemical  equation  and  table  of 
combining  weights,  it  is  possible  to  state  what  relative 
weight  of  each  element  is  concerned  in  any  chemical  action, 
it  must  never  be  forgotten  that  such  weights  are  always 
determined  by  chemical  analysis,  and  that  from  the  data 
thus  obtained  the  equation  is  compiled.  So,  too,  the  series 
of  combining  weights  is  simply  a  tabular  expression  of  results 
obtained  by  practical  analytic  investigation. 

8 1 .  Calculation  of  Percentage  Composition. — Know- 
ing the  molecular  weight  of  any  substance,  the  number  of 
atoms  it  contains,  and  the  atomic  weight  of  each  constituent 
atom,  we  can  easily  find  the  percentage  composition  (that  is, 
the  composition  of  100  parts  of  a .  substance,  the  form  in 
which  the  results  of  analyses  are  usually  expressed). 

Let        /#  =  molecular  weight; 
a  =  atomic  weight ; 
n  =  number  of  atoms  of  a  constituent; 
x  =  percentage. 


§  5  THEORETICAL  CHEMISTRY  65 

Then,  we  have  the  proportion 

;;/  :  an  =  100  :  x 
From  this  proportion  we  obtain  the  formula 

_  100  an  (1  } 

~m~  (    •> 

Case  L  —  To  find,  therefore,  the  percentage  composition  of 
a  molecule  when  the  molecular  weight,  number  of  atoms, 
and  atomic  weights  are  known,  use  the  following  rule  : 

Rule.  —  Multiply  tJie  atomic  weight  by  the  number  of  atoms, 
and  the  product  by  100;  divide  by  the  molecular  ivcigJit,  and 
the  quotient  will  be  the  percentage  amount  of  that  con- 
stituent. 

EXAMPLE.  —  The  formula  for  potassium  chlorate  is  KCIO^\  assigning 
to  each  atom  its  weight,  we  have  the  equation 

K  +    Cl    +    O3     =    KC/0, 

39  35.5         (16x3)  122.5 

which  means  that  122.5,  being  the  sum  of  the  constituent  atoms,  repre- 
sents the  molecular  weight  of  potassium  chlorate,  and  that  122.5  parts 
of  KCIO-A  contain  48  parts  of  oxygen,  39  parts  of  potassium,  and  35.5 
parts  of  chlorine.  What  quantities  of  these  constituents  will  be  found 
in  100  parts  of  KCIOJ. 
SOLUTION.  — 


(1)  Oxygen;          x=  =  39.  18g  of  oxygen.     Ans. 

\&&.  o 

(2)  Chlorine;        x  =    10°  *8'5;5  X  1  =  28.98^  of  chlorine.     Ans. 

I**,  o 

(3)  Potassium;     x=       10°  *3*?  X  *  =  31.  84g  of  potassium.     Ans. 

\ii£.  5  .  - 

Total,  100.  00£. 

82.  Other  Problems  by  This  Formula.  —  In  formula  1 
the  four  quantities  «,  ;/,  /;/,  and  x  are  employed.  Any  three 
of  these  being  known,  the  fourth  can,  of  course,  be  easily 
found.  There  are,  consequently,  three  more  cases  to  be 
considered. 

Case  II.  —  To  find  the  number  of  atoms  of  the  constituents 
of  a  compound,  that  is,  the  chemical  formula  of  the  molecule, 
having  given  the  percentage  amount  of  the  constituents,  their 


6(5  THEORETICAL  CHEMISTRY.  §  5 

atomic  weights  and  the  molecular  weight  of  the  compound, 
formula  1,  by  transposition,  becomes 

n  =   mx ,  (2.) 

from  which  we  derive  the  following  rule: 

Rule. — Multiply  the  molecular  weight  by  the  percentage  of 
the  given  constituent,  and  divide  the  product  by  the  atomic 
iv eight  of  the  constituent,  multiplied  by  100;  the  quotient  is 
the  number  of  atoms  of  that  constituent  in  the  molecule. 

Ex  AMPLE.  ^Sulphuric  acid  has  a  molecular  weight  of  98;  its  percent- 
age composition  is:  hydrogen  2.04  per  cent.,  sulphur  32.65  per  cent. , 
and  oxygen  65.31  per  cent.  The  atomic  weight  of  hydrogen  is  1,  that 
of  sulphur  32,  and  that  of  oxygen  16.  What  is  the  formula  of  sulphuric 
acid? 

SOLUTION. — The  atomic  weights  of  hydrogen,  sulphur,  and  oxygen, 
are,  respectively,  1,  32,  and  16;  the  number  of  atoms  of  each  element 
is  obtained  by  formula  2. 
98X2.04 


IX  100 
98x32.65 

32  X  100 
98X65.31 


=  2,  number  of  atoms  of  hydrogen 
—  1,  number  of  atoms  of  sulphur; 
=  4,  number  of  atoms  of  oxygen. 


16  X  100 
The  molecular  formula  of  sulphuric  acid  is,  therefore,  H^SO*.     Ans. 

Case  ///. — Having  the  molecular  weight,  the  percentage 
composition,  and  the  number  of  atoms  of  any  constituent  in 
the  molecule,  to  find  the  atomic  weight  of  that  constituent, 
we  obtain  from  formula  1,  by  transposition, 

mx  ,Q  \ 

=  looT?         <8<> 

from  which  we  have  the  following  rule : 

Rule. — Multiply  the  molecular  weiglit  by  the  percentage  of 
the  constituent  whose  weight  is  desired,  and  divide  the  prod- 
uct by  the  number  of  atoms  multiplied  by  100;  the  quotient  is 
the  atomic  weight  required. 

EXAMPLE. — Mercurous  chloride  contains  2  atoms  of  mercury,  consti- 
tuting 84.96  per  cent,  of  the  whole;  the  molecular  weight  of  mercurous 
chloride  being  470.34,  what  is  the  atomic  weight  of  mercury? 


§  5  THEORETICAL  CHEMISTRY.  07 

SOLUTION.  —  Substituting  these  values  in  formula  3,  we  have 
a  —  -  2  V  100  —  =  •^•^'  the  atomic  weight  of  mercury.     Ans. 

Case  IV.  —  Having  given  the  atomic  weight  of  any  con- 
stituent, its  percentage,  and  the  number  of  atoms  of  it  in  the 
molecule,  to  find  the  molecular  weight  of  the  compound,  we 
obtain,  by  a  final  transposition  of  formula  1, 

WOan 
m  =  ——.  (4.) 

Whence,  the  following  rule  : 

Rule.  —  Multiply  the  atomic  weight  of  the  given  constituent 
by  the  number  of  its  atoms,  and  this  product  by  100;  divide 
the  final  product  by  the  percentage  of  that  constituent,  and 
the  quotient  is  the  molecular  weight. 

EXAMPLE.  —  One  molecule  of  sulphuric  acid  contains  1  atom  of  sul- 
phur, which  is  32.65  per  cent,  of  it;  the  atomic  weight  of  sulphur  being 
32,  what  is  the  molecular  weight  of  sulphuric  acid  ? 

SOLUTION.  —  Substituting  these  values  in  formula  4, 

m  =    ^  -  •  =  98,  the  molecular  weight  of  sulphuric  acid.     Ans. 

o^.  bo 

83.  Calculation  of  an  Atomic  Group.  —  It  some- 
times becomes  necessary  to  calculate  the  percentage  of  a 
group  of  atoms  in  a  molecule.  Formula  1  enables  us  to  do 
this,  using  a  to  indicate  the  weight  of  the  group,  and  n  the 
number  of  such  groups  in  the  molecule. 

EXAMPLE.  —  The  mineral  magnesite  MgCO*  is  decomposed  by  heat 
into  MgO  and.Q92;  how  much  magnesium  oxide  MgO  is  there  in  100 
parts  of  magnesite  ? 

SOLUTION.—  The  molecular  weight  of  MgO  is  40,  and  of  MgCO*  is  84. 

a  n  X  100 

By  formula  1,  -  ,  we  have 
m 


40  X  =  47.62#  of  magnesium  oxide.     Ans. 


84.  Other  Than  Percentage  lumbers.  —  In  some  cases 
it  is  necessary  to  calculate  the  quantity  of  a  constituent  in 
more  or  less  than  100  parts.  The  answers  to  such  problems 
can,  of  course,  be  obtained  by  stating  the  proportion  for 


68  THEORETICAL  CHEMISTRY.  §  5 

each  problem  ;  but  they  may  also  be  derived  from  formula  1 
by  putting  y,  the  quantity  of  the  constituent,  in  place  of  x, 
and  s,  the  quantity  of  the  compound,  in  place  of  100.  The 
formula  then  becomes 


from  which  we  obtain  the  following  rule  : 

Rule.  —  Multiply  the  weight  of  the  constituent  contained  in 
one  molecule  by  the  weight  of  the  compound  given  in  "the  prob- 
lem, and  divide  this  product  by  the  molecular  weight;  the 
quotient  will  be  the  weight  of  the  required  constituent. 

EXAMPLE.  —  How  much  iodine  may  be  obtained  from  311  grams  of 
potassium  iodide  KI,  the  atomic  weight  of  iodine  being  127,  and  the 
molecular  weight  of  potassium  iodide  being  166  ? 

SOLUTION.  —  Substituting  the  proper  values  in  formula  5,  we  obtain 

' 


Hence,  311  grams  of  potassium  iodide  yield  237.9  grams  of  iodine.    Ans. 

Finally,  if  we  wish  to  know  the  quantity  of  a  compound 
necessary  to  yield  a  certain  weight  of  the  constituent,  we 
obtain  by  transposition  the  following  formula : 

(6.) 


a  n 

EXAMPLE.— How  much  potassium  iodide  would  be  required  to  yield 
59  grams  of  iodine  ? 

SOLUTION. — Substituting  the  proper  values  in  formula  6, 

2  =  • — — = —  =  77.12  grams  of  potassium  iodide.     Ans. 


85.  Calculations  From  Equations. — The  same  simple 
arithmetical  principles  are  applied  in  calculating  the  weights 
of  substances  that  enter  into  or  are  produced  by  chemical 
changes.  The  change  should  be  expressed  by  an  equation. 
The  molecular  weights  of  all  the  participating  compounds 
are  calculated  from  a  table  of  atomic  weights  and  are  placed 
below  their  formulas.  Having  got  thus  far,  the  problems 
may  be  solved  by  making  the  following  proportion : 


§  o  THEORETICAL  CHEMISTRY.  69 

The  molecular  weight  of  the  given  substance  is  to  the  abso- 
lute weight  of  it  given  in  the  problem,  as  the  molecular  weight 
of  the  required  substance  is  to  its  required  quantity. 

Let       M  =  molecular  weight  of  the  given  substance; 

;;/    —  molecular  weight  of  the  required  substance  ; 
W  —  absolute  weight  of  the  given  substance  ; 
w   =  absolute  weight  of  the  required  substance. 

We  have  the  proportion  M  :  W  =  ;;/  :  w,  whence  we  obtain 
the  following  formulas  : 

M  ---.  ^.          (7.) 

(8.) 
m 

Mw 

m  =  ~W~'         ^  ') 

»=».          (10.)  - 


Hence,  if  we  know  any  three  of  the  above  values 
(M,  W,  m,  and  w),  we  can  find  the  fourth  by  using  one  of 
the  above  formulas.  The  application  of  these  formulas  to 
the  four  possible  cases  is  shown  by  the  following  examples  : 

Case  I.  —  Knowing  the  absolute  and  molecular  weights  of 
the  given  substance,  and  also  the  absolute  weight  of  the 
required  substance,  to  find  the  molecular  weight  of  the 
required  substance.  > 

EXAMPLE.—  A  certain  chemical  reaction  is  represented  by  the  follow- 
ing equation: 

Zn(NO»)3    +    A"2C<93     =    ZnCO3    +    (KNO& 

zinc  potassium  zinc  potassium 

nitrate  carbonate         carbonate  nitrate 

189  138  125  202 

If  156  grams  of  zinc  nitrate  yield  103.  17  grams  of  zinc  carbonate,  and 
the  molecular  weight  of  zinc  carbonate  is  125,  what  is  the  molecular 
weight  of  zinc  nitrate  ? 

SOLUTION.  —  Substituting  the  values  just  given  in  formula  7,  we  obtain 

M  =         *      '    =  189,  the  molecular  weight  of  zinc  nitrate.     Ans. 
10o.  17 


70  THEORETICAL  CHEMISTRY.  §  5 

Case  II.  —  Knowing  the  molecular  weight  of  both  the  given 
and  also  the  required  substances,  to  find  the  absolute  weight 
of  the  given  substance  necessary  to  yield  a  certain  weight  of 
the  required  substance. 

EXAMPLE.  —  Sodium  nitrate  is  prepared  by  the  action  of  hydro- 
disodium  phosphate  upon  strontium  nitrate,  according  to  the  following 
equation  : 

Oi      =      HSrPO, 


strontium  hydro-di-  hydrostrontium  sodium 

nitrate  sodium  phosphate  phosphate  nitrate 

211  142  183  170 

How  much  hydro-disodium  phosphate  will  be  required  in  order  to 
yield  809  grams  of  sodium  nitrate  ? 

SOLUTION.  —  Substituting  the  proper  values  in  formula  8,  we  obtain 

142  V  309 
W  =  —  ^=Q  -  =  258.  1  grams  of  hydro-disodium  phosphate.     Ans. 

Case  III.  —  Knowing  the  absolute  and  molecular  weights 
of  the  given  substance  and  also  the  absolute  weight  of  the 
required  substance,  to  find  the  molecular  weight  of  the  latter. 

EXAMPLE.  —  Sodium  carbonate  is  produced  by  the  action  of  sodium 
sulphide  upon  calcium  carbonate  according  to  the  following  equation  : 

Na^S  +     CaCO3  =    Na^CO3  +     CaS 

sodium  calcium  sodium  calcium 

sulphide         carbonate  carbonate         sulphide 

78  100  106  72 

103.17  grams  of  sodium  carbonate  are  obtained  from  97.33  grams  of 
calcium  carbonate  ;  the  molecular  weight  of  calcium  carbonate  being 
100,  what  is  the  molecular  weight  of  sodium  carbonate  ? 

SOLUTION.  —  Substituting  these  values  in  formula  9,  we  obtain 
100X103.17 


97.33 


=  106,  the  molecular  weight  of  sodium  carbonate.  Ans. 


Case  IV.  —  Knowing  the  molecular  weights  of  both  the 
given  and  the  required  substance,  and  also  the  absolute 
weight  of  the  given  substance,  to  find  the  absolute  weight 
of  the  required  substance. 

EXAMPLE.  —  Hydropotassium  sulphate  is  produced  by  the  action  of 
sulphuric  acid  upon  potassium  nitrate,  according  to  the  following 
equation  : 

+    H*SO<    =    H(NO,}    +    HKSO, 


potassium  sulphuric  nitric  hydropotassium 

nitrate  acid  acid  sulphate 

101  98  63  136 


§  5  THEORETICAL  CHEMISTRY.  71 

How  much  hydropotassium  sulphate  is  yielded  by  the  decomposition 
of  500  grams  of  potassium  nitrate  by  sulphuric  acid  ? 

SOLUTION. — Substituting  the  proper  values  in  formula  1O,  we  obtain 

500  V  136 
w  =  - — rrr: =  673.27  grams  of  hydropotassium  sulphate.     Ans. 

In  all  the  above  problems  it  has  been  assumed  that  each 
molecule  of  the  factor  yielded  one  of  the  product.  If  in  any 
reaction  this  is  not  true,  then  J/and  m  must  represent  the 
sum  of  the  molecular  weights  expressed  in  the  equation. 

86.  As  all  gaseous  molecules  have  the  same  volume,  we 
are  able  to  read  every  equation  representing  a  reaction 
between  gaseous  bodies,  not  only  quantitatively  but  also 
volume  trically. 

Thus,  the  equation 

2//2  +   <92  =  2//26> 
4  32  36 

shows  that  2  molecules  of  hydrogen  and  1  molecule  of  oxygen 
yield  2  molecules  of  water.  It  may  also  be  read:  2  volumes 
of  hydrogen  and  1  volume  of  oxygen  yield  2  volumes  of 
water,  and  calculations  may  be  made  from  this  as  well  as 
from  the  weights. 

EXAMPLE. — How  much  water  would  be  yielded  by  the  explosion  of 
10  cubic  centimeters  of  hydrogen  ? 

SOLUTION. — Since  2  volumes  of  hydrogen  yield  2  volumes  of  water, 
10  cubic  centimeters  of  hydrogen  will  yield  10  cubic  centimeters  of 
water  in  the  gaseous  state — steam. 

8  7 .     Relation  of  the  Hydrogen  Unit  to  the  Air  Unit.— 

Density  has  .already  been  denned  as  the  weight  of  any 
volume  of  gas  compared  with  that  of  the  same  volume  of 
hydrogen,  measured  at  the  same  temperature  and  pressure. 
(These  conditions  are  always  understood  in  speaking  of 
comparative  weights  of  gases.)  The  term  specific  gravity 
may  likewise  be  defined  as  the  weight  of  any  volume  of  gas 
compared  with  that  of  the  same  volume  of  air.  The  density 
of  hydrogen  is  assumed  to  be  1,  and  its  specific  gravity  is 
.0693.  It  is  then  obvious  that  by  multiplying  the  density  of 
any  gas  by  the  specific  gravity  of  hydrogen,  i.  e.,  .0693,  the 
specific  gravity  of  the  gas  in  question  may  be  obtained,  and 


72  THEORETICAL  CHEMISTRY.  §  5 

that,  conversely,  by  dividing  the  specific  gravity  of  any  gas 
by  .0693  its  density  may  be  found. 

ILLUSTRATION. — The  density  of  oxygen  is  16;  its  specific  gravity, 
therefore,  will  be  16  X  -0693  =  1.108. 

The  specific  gravity  of  chlorine  is  2.46;  its  density,  therefore,  is 
2.46-T-.0693  =  35.5. 


PRESSURE,  VOLUME,  AND  TEMPERATURE  OF  GASES. 

88.  If  the  temperature  of  a  confined  gas  remains  the 
same,  the  pressure  and  volume  will  always  vary  according 
to  Mariotte's  law,  which  is  as  follows  : 

The  temperature  remaining  the  same,  the  quantity  .of  gas 
varies  inversely  as  the  pressure. 

The  meaning  of  this  is  :  If  the  volume  of  the  gas  is  dimin- 
ished to  \,  \,  \,  etc.  of  its  former  volume,  the  pressure  will 
be  increased  2,  3,  5,  etc.  times,  or  if  the  outside  pressure  is 
increased  2,  3,  5,  etc.  times,  the  volume  of  the  gas  will  be 
diminished  to  \,  \,  \,  etc.  of  its  original  volume,  the  temper- 
ature remaining  constant. 

As  a  necessary  consequence  of  Mariotte's  law,  it  may  be 
stated  that  the  density  of  a  gas  varies  directly  as  the  pressure, 
and  inversely  as  the  volume  ;  that  is,  the  density  increases  as 
the  pressure  increases,  and  deer  eases,  as  the  volume  increases. 

Hence,  the  volume  of  gases  changes  with  the  variation  of 
atmospheric  pressure  as  measured  by  the  barometer;  that  is, 
the  volume  is  increased  when  the  barometer  falls,  and 
decreased  as  the  barometer  rises.  The  normal  pressure  to 
which  it  is  usual  to  refer  gaseous  volumes  is  760  millimeters 
of  mercury. 

If  we  represent  the  volume  of  a  gas  under  a  height  H  of 
the  barometer  by  V,  and  under  any  other  height  h  by  v, 
then,  according  to  the  above  law,  we  obtain  the  proportion  : 

V  :  v  rr  h  :  H, 
which  is  expressed  by  the  formula  : 


Rule.  —  Multiply  the  given  volume  of  the  gas  by  the  given 


§  5  THEORETICAL  CHEMISTRY.  73 

height  of  the  barometer  ;  divide  t/te  product  by  !t>0,  and  the 
quotient  tlius  obtained  will  be  the  normal  volume. 

89.  Mariotte's  law  holds  when  the  temperature  of  the 
gas  remains  constant;  if  the  temperature  varies,  for  every 
increase  of  temperature  there  will  be  a  corresponding 
increase  of  volume.  The  law  that  expresses  this  change  is 
called  Gay-Lussac's  law,  and  may  be  stated  as  follows : 

If  the  pressure  remains  constant,  every  increase  of  temper- 
ature of  1°  produces  in  a  given  quantity  of  gas  an  expansion 
of  ^\^  of  its  volume  at  0°. 

If  the  pressure  remains  constant,  it  will  also  be  found  that 
every  decrease  of  temperature  of  1°  will  cause  a  decrease 
of  24¥  of  the  volume  at  0°. 

Let     v  =  original  volume  of  gas; 
i\  =  final  volume  of  gas ; 

/   —  temperature  corresponding  to  volume  v; 
tl  =  temperature  corresponding xto  volume  vv. 

Then, 

Rule. —  The  volume  of  gas  after  heating  (or  cooling)  equals 
the  original  volume  multiplied  by  273  plus  the  final  tempera- 
ture, divided  by  273  plus  the  original  temperature. 

EXAMPLE  1. — A  gas  measures  30  cubic  centimeters  atO°;  what  will 
it  measure  at  72°? 

SOLUTION. — Substituting  in  formula  12  the  proper  values,  we  obtain 

.^i 

EXAMPLES. — A  certain  quantity  of  gas  measures  36.6  cubic  centi-. 
meters  at  96° ;  what  will  it  measure  at  0°  ? 

SOLUTION. — Substituting  in  formula  12  the  proper  values,  we  obtain 


(O*7Q    i   7O\ 
g73^  Q  )  -  37.912  c.  c.     Ans. 


*"  =  36'6  =  27'08  c'  c'    Ans' 


EXAMPLE  3.  —  A  certain  quantity  of  oxygen  measures  560  cubic  centi- 
meters at  15°  ;  what  will  the  same  quantity  measure  at  95°  ? 

SOLUTION.  —  Substituting  in  formula  12  the  proper  values,  we  obtain 

=  715.55  c.  c.     Ans. 


74  THEORETICAL  CHEMISTRY.  §  5 

EXAMPLES  FOR  PRACTICE. 

90.  Solve  the  following : 

1.  The  formula  of  calcium  chloride  is  CVzC/2;  its  molecular  weight 
is  111.     What  percentage  quantities  of  the  two  constituents  will  be 
found  in  100  parts  of  calcium  chloride  ?  »     ,    j  36$  of  calcium. 

(  64$  of  chlorine. 

2.  Potassium  chlorate  has  a  molecular  weight  of  122.5,  and  is  com- 
posed of  89.18  per  cent,  of  oxygen,  28.98  per  cent,  of  chlorine,  and 
31.84  per  cent,  of  potassium.     What  is  the  formula  of  potassium  chlo- 
rate ?  Ans.  KCIO*. 

3.  Salt  NaCl  contains  39.32  per  cent,  of  sodium,  whose  atomic 
weight  is  23.     In  a  molecule  of  salt  there  is  but  1  atom  of  sodium. 
What  is  the  molecular  weight  of  salt?  Ans.  58.5. 

4.  A  quantity  of  ammonia  gas  had  a  volume  of  350  cubic  centimeters 
when  measured  at  74° ;  what  volume  would  it  have  at  0°? 

Ans.  275.3  c.  c. 

5.  The  molecular  weight  of  silver  nitrate  AgNO*  is  170 ;  it  contains 
63. 53  per  cent,  of  silver,  and  has  but  1  atom  of  silver  in  a  molecule. 
What  is  the  atomic  weight  of  silver?  Ans.   108. 

6.  Nitric  acid  is  prepared  by  the  action  of  sulphuric  acid  on  potas- 
sium nitrate,  according  to  the  equation : 

KNO*    +    H^SO,    =    HNO*    +    HKSO, 

potassium         sulphuric  nitric          hydropotassium 

nitrate  acid  acid  sulphate 

101  98  63  136 

(a)  How  much  potassium  nitrate  is  necessary  to  yield  36  grams  of 
nitric  acid?  Ans.  57.71  grams. 

(b)  How   much   nitric   acid   may  be   produced  from  500  grams   of 
potassium  nitrate  ?  Ans.  311.88  grams. 

(c}     How  much  sulphuric  acid  will  be  required  to  yield  36  grams  of 
nitric  acid  ?  Ans.  56  grams. 

7.  The  density  of  marsh  gas  C//4  is  8;  how  many  grams  does  a 
liter  of  it  weigh  ?  •  Ans.  .7168  gram. 

8.  The   specific   gravity  of   hydrogen   iodide  is  4,41;     what  is  its 
molecular  weight  ?  Ans.   127.26. 

CRYSTALLOGRAPHY. 

91.  Most  chemical  substances,  when  they  pass  from  the 
liquid  or  gaseous  into  the  solid  stato,  assume  some  definite 
geometric  form,  or  are  said  to  crystallize.      Crystals  are  pro- 
duced when  a  substance,  such  as  nitrate  of  potash,  is  dis- 
solved  in  water   and  the  solution  is  allowed  to  gradually 


§  5  THEORETICAL  CHEMISTRY.  75 

evaporate;  when  a  body,  such  as  sulphur,  is  melted  and 
allowed  to  solidify  by  cooling-;  or  when  a  volatile  substance, 
such  as  iodine,  is  vaporized,  and  the  vapor  condensed  on  a 
cool  surface. 

One  of  the  most  common  methods  of  crystallizing-  a  solid 
substance  consists  in  dissolving  it  in  hot  water  and  allowing 
the  solution  to  cool  slowly.  The  more  slowly  it  cools,  the 
larger  and  more  symmetrical  are  the  crystals. 

A  hot  saturated  solution  is  not  generally  the  best  for  crys- 
tallizing, because  it  deposits  the  dissolved  body  too  rapidly. 
Thus,  the  hot  solution  of  saltpeter  prepared  as  above  would 
solidify  to  a  mass  of  minute  crystals  on  cooling,  but  if  100 
grams  of  saltpeter  are  dissolved  in  120  cubic  centimeters  or 
4  measured  ounces  of  boiling  water,  the  solution  will  form 
crystals  2  or  3  inches  long  when  slowly  cooled  in  a  covered 
vessel.  If  the  solution  is  stirred  while  cooling,  the  crystals  will 
be  very  minute,  having  the  appearance  of  a  white  powder. 

Some  solids,  however,  refuse  to  crystallize,  even  from  a 
hot  saturated  solution,  if  it  is  kept  absolutely  undisturbed. 
Sodium  vSulphate  affords  a  good  example  of  this.  If  to  boil- 
ing water  in  a  flask  the  sulphate  is  added  until  the  water 
refuses  any  longer  to  take  it  up,  it  will  be  found  that  the 
water  has  dissolved  more  than  twice  its  weight  of  the  salt, 
and  that  a  solution  results  which  boils  at  104.5°.  If  this 
solution  is  allowed  to  cool  in  the  open  flask,  an  abundant  crys- 
tallization will  take  place,  for  cold  water  will  dissolve  only 
about  one-third  of  its  weight  of  sulphate.  But  if  the  flask  is 
tightly  corked  while  the  solution  is  boiling,  it  may  be  kept 
for  several  days  without  crystallizing.  In  this  condition  we 
say  the  solution  is  supersaturated.  On  withdrawing  the 
cork,  the  air  entering  the  partly  vacuous  space  above  the 
liquid  will  be  seen  to  disturb  the  surface  slightly,  and  from 
that  point  beautiful  prisms  will  shoot  through  the  liquid, 
until  the  whole  has  become  a  nearly  solid  crystalline  mass. 
A  considerable  elevation  of  temperature  is  observed,  conse- 
quent on  the  passage  from  the  liquid  to  the  solid  form.  If 
the  solution  of  sodium  sulphate  is  somewhat  weaker,  con- 
taining exactly  two-thirds  of  its  weight  of  the  sulphate,  it 


76  THEORETICAL  CHEMISTRY.  §  5 

may  be  cooled  without  crystallizing1,  but  a  touch  with  a  glass 
rod  will  start  the  crystallization  immediately. 


The  crystallization  of  a  supersaturated  solution  is 
caused  by  contact  with  a  crystal  of  the  salt  itself.  Minute 
crystals  of  sodium  sulphate  are  present  in  the  floating  dust 
of  the  air  and  cause  the  crystallization  wrhen  they  fall  into 
the  supersaturated  solution.  A  perfectly  clean  glass  rod 
may  be  dipped  into  the  solution  without  causing  crystalliza- 
tion, but  a  rod  that  has  been  exposed  to  air  will  have  some 
particles  of  sodium  sulphate  on  it  and  will  start  crystalliza- 
tion ;  if  the  rod  is  heated  so  as  to  render  the  sodium  sulphate 
from  the  dust  anhydrous,  it  will  no  longer  cause  crystalliza- 
tion unless  it  is  drawn  through  the  hand. 

Air  filtered  through  cotton  wool  does  not  cause  super- 
saturated solutions  to  crystallize.  If  the  solution  contain- 
ing- two-thirds  of  its  wreight  of  sodium  sulphate  is  allowed  to 
cool  in  a  flask  closed  by  a  cork  furnished  with  two  tubes 
plugged  with  cotton  wool,  it  will  be  found  that  on  with- 
drawing the  plugs  and  blowing  through  one  of  the  tubes 
dipping  into  the  solution,  crystallization  does  not  take  place  ; 
but  if  air  is  blown  by  a  pair  of  bellows  into  the  same  solution 
it  will  crystallize  at  once. 

A  striking  illustration  of  the  power  of  unfiltered  air  to 
start  crystallization  is  afforded  by  a  solution  of  alum  pre- 
pared by  saturating  a  volume  of  water  at  90°  and  allowing 
it  to  cool  in  a  flask,  the  mouth  of  which  is  closed  by  a  plug 
of  cotton  wool.  In  this  state  it  may  be  kept  for  weeks  with- 
out crystallizing,  but  on  withdrawing  the  plug,  crystallization 
will  be  seen  to  commence  at  once  at  a  few  points  on  the  sur- 
face immediately  under  the  opening  of  the  neck,  and  will 
spread  slowly  from  these,  octahedrons  of  alum  of  half  an 
inch  or  more  in  diameter  being  built  up  in  a  few  seconds, 
the  temperature,  at  the  same  time,  rising  very  considerably. 

In  the  laboratory,  stirring  is  always  resorted  to  in  order  to 
induce  crystallization,  if  it  does  not  take  place  immediately. 
Thms,  it  is  usual  to  test  for  potassium  in  a  solution  by  adding 
tartaric  acid,  which  should  cause  the  formation  of  minute 


§  5  THEORETICAL  CHEMLSTRY.  77 

crystals  of  hydropotassium  tartrate  (cream  of  tartar),  but  the 
test  seldom  succeeds  unless  the  solutions  are  stirred  briskly 
with  a  glass  rod. 

93.  The  crystals  of  sodium  sulphate  produced  in  the 
experiment  described  above  contain,  in  a  state  of  combina- 
tion with  the  salt,  more  than  half  their  weight  of  water. 
Their  composition  is 

anhydrous  sodium  sulphate  JVa9SOt,  142  parts,  or  1  molecule, 
water      .       ...       .       .         H^O,         180  parts,  or  10  molecules, 

as  expressed  by  the  formula  Na^SO^Hfl,  or  Na^SO^ 
Wag  (aq  being  the  abbreviation  of  aqua,  the  Latin  name  for 
water).  If  some  of  the  crystals  are  pressed  between  blotting 
paper  to  remove  adhering  water,  and  left  exposed  to  air, 
they  will  gradually  effloresce,  or  become  covered  with  a  white, 
opaque  powder.  This  powder  is  the  anhydrous  sodium 
sulphate  into  which  the  entire  crystals  ultimately  become 
converted  by  exposure  to  air,  since  most  crystals  containing 
water  have  their  crystalline  forms  destroyed  by  the  loss  of  the 
water,  which  is  commonly  spoken  of  as  water  of  crystallization. 
Colored  salts  containing  water  of  crystallization  generally 
change  color  when  the  water  is  removed.  The  sulphate  of 
copper  affords  a  very  good  example  to  this.  The  beautiful 
blue  prismatic  crystals  of  this  salt  contain 

anhydrous  sulphate  of  copper  CuSO±,  159.5  parts,  or  1  molecule, 
water      .       .       .       .       .  .    .     //aO,        90     parts,  or  5  molecules, 

which  is  expressed  by  the  formula  CuSOjbHjD. 

When  these  are  exposed  to  the  air  at  the  ordinary  temper- 
ature they  remain  unchanged ;  but  if  heated  to  the  boiling 
point  of  water  they  become  opaque  and  are  easily  crumbled 
into  a  grayish-white  powder.  This  powder  consists  of 

anhydrous  sulphate  of  copper  CuSO4,  159.5  parts,  or  1  molecule, 
water H<iO,        18     parts,  or  1  molecule, 

as  expressed  by  the  formula  CuSO^H^O.  The  four  water 
molecules  that  have  been  removed  by  the  heat  constituted 
the  water  of  crystallization,  on  which  the  form  and  color  of 
the  sulphate  of  copper  depend.  Moistening  this  white 
powder  with  a  little  water,  the  blue  color  is  restored  with 


78  THEORETICAL  CHEMISTRY.  §  5 

the  evolution  of  a  considerable  amount  of  heat.  The  one 
molecule  of  water  that  remained  after  the  application  of 
heat  cannot  be  removed  unless  the  salt  is  heated  to  199°, 
which  undoubtedly  proves  that  it  is  held  to  the  copper 
sulphate  by  a  more  powerful  attraction.  On  this  account  it 
is  spoken  of  as  water  of  constitution,  and  in  order  that  the 
formula  of  the  salt  may  exhibit  the  difference  between  the 
water  of  constitution  and  that  of  crystallization,  it  is  some- 
times written  CuSO  ^,Hfl,kaq. 

We  may  now  define  water  of  crystallization  as  that  water 
which  is  generally  expelled  from  a  crystal  at  100°,  and  which 
is  closely  allied  with  the  form  and  color  of  the  crystal. 
Water  of  constitution  is  not  generally  expelled  at  100°,  and 
is  in  more  intimate  connection  with  the  chemical  properties 
of  the  compound  forming  the  crystal. 

94.  Many  native  minerals  exhibit  perfect  crystalline 
forms;  though  our  investigations  have  not  yet  disclosed  the 
process  of  their  formation.  Judging  from  their  perfect 
shape  we  are  justified  in  assuming  that  the  process  of  their 
formation  must  have  been  a  necessarily  slow  one.  Besides 
their  regular  shape,  these  crystalline  bodies  possess  other 
remarkable  properties,  that  is,  a  peculiar  power  of  splitting 
in  certain  directions  more  readily  than  in  others,  a  property 
that  is  known  as  cleavage;  and  in  many  cases  they  possess 
the  property  of  allowing  light  to  pass  more  readily  in  certain 
directions  than  in  others,  giving  rise  to  the  well  known 
phenomena  of  double  refraction  (see  Art.  158,  Physics). 

A  body  assuming  two  distinct  crystalline  forms  is  said  to 
be  dimorphous;  the  study  of  even  the  more  familiar  ele- 
ments includes  some  very  important  instances  of  dimorphism. 
A  body  that  does  not  occur  in  crystals  is  termed  amorpJious; 
i.  e. ,  without  crystalline  form.  Certain  complicated  struc- 
tures of  the  vegetable  and  animal  world  exhibit  a  structure 
that,  although  non-crystalline,  is  not  devoid  of  a  certain 
definite  arrangement,  and  to  which  the  name  organized,  or 
cellular,  structure  has  been  given. 

As  a  rule,  every  particular  substance  possesses  a  definite 


THEORETICAL  CHEMISTRY. 


79 


form,  in  which  it  will  always  crystallize,  and  by  which  it  may 
be  distinguished ;  when  a  crystal  is  formed  from  aqueous  solu- 
tion, for  instance,  the  most  minute  particle  possesses  the  same 
crystalline  structure  and  perfect  form  as  the  largest  crystals. 
Certain  substances  exhibiting  a  similarity  in  their  chemical 
constitution  are  found  to  crystallize  in  the  same  forms; 
these  are  said  to  be  isomorphovs. 

95.  Science  has  succeeded  in  classifying  the  thousands 
of  known  crystals  in  six  systems,  to  each  of  which  belongs  a 
number  of  forms  having  some  property  in  common.  In 
order  to  classify  these  different  crystals,  the  existence  of 
certain  lines  within  the  crystal,  called  axes,  is  assumed, 
around  which  the  form  can  be  symmetrically  built  up. 
These  axes  are  assumed  to  intersect  in  the  center  of  the 
crystal,  and  to  pass  through  from  one  side  to  the  other. 

First,  or  Regular,  System. — Three  axes,  all  equal  and 
at  right  angles  to  each  other.  The  simple  forms  of  this 
system  are  the  cube,  Fig.  G;  the  regular  octahedron,  Fig.  7; 
the  rhombic  dodecahedron,  Fig.  8 ;  and  the  regular  tetrahe- 
dron, Fig.  9. 


FIG.  6.  FIG.  7.  FIG.  8.  FIG. 

The  following  are  examples  of  substances  crystallizing  in 
this  system:  common  salt,  diamond,  alum,  iron  pyrites, 
garnet,  fluorspar,  etc. 

Second,  or  Quadratic,  System. — Three  axes,  all  at  right 
angles,  one  either  shorter  or  longer  than  the  other  two. 


FIG.  10. 


FIG.  11. 


FIG. 


80 


THEORETICAL  CHEMISTRY. 


The  simple  forms  of  this  system  are  the  first  right  square 
prism,  Fig.  10;  the  second  right  square  prism,  Fig.  11;  the 
first  right  square  octahedron,  Fig.  12;  and  the  second  right 
square  octahedron,  Fig.  13. 

In  the  first  square  prism,  the  axes 
terminate  in  the  center  of  each  of 
.the  sides,  and  in  the  second  square 
prism,  the  axes  terminate  at  the 
intersection  of  the  sides,  but  this 
is  reversed  with  regard  to  the 
octahedron. 

The  following  are  examples  of 
substances  crystallizing  in  this  system:  stannic  oxide, copper 
pyrites,  yellow  prussiate  of  potash,  etc. 

Third,  or  Rhombic,  System. — Three  axes,  all  unequal, 
and  all  at  right  angles.  The  chief  forms  of  the  crystals  in 
this  system  are  the  right  octahedron  with  rhombic  base. 
Figs.  14  and  15,  and  the  right  rliombic  prism,  Fig.  16. 


FIG.  1 


FIG.  14. 


FIG.  15. 


FIG.  16. 


Examples  of  this  class  are :  native  sulphur,  calcium  sulphate, 
barium  sulphate,  magnesium  sulphate,  calcium  carbonate, 
potassium  nitrate,  etc. 

Fourth,  or  Monoclinic,  System. — Three  axes,  all  unequal ; 
two  cut  each  other  obliquely,  and  one  is  at 
right  angles  to  the  plane  of  the  other  two. 
The  oblique  rhombic  octahedron,  Fig.  17,  be- 
longs to  this  system. 

Examples  of  this  class  are:  sodium  car- 
bonate, sodium  phosphate,  borax,  cane  sugar, 
ferrous  sulphate,  etc. 

Fifth,  or  Triclinic,  System. — Three  axes,  all  unequal, 
and  all  oblique.  The  doubly  oblique  octaJiedron,  and  the 
doubly  oblique  prism,  Fig.  18,  are  the  leading  forms  of  this 


FIG.  17. 


§  5  THEORETICAL  CHEMISTRY.  81 

system.       Copper   sulphate    CuSO^5Hfl,    boric   acid,    the 
mineral  albite,  potassium  bichromate,  and  a  few  other  sub- 


FlG.  18.  FIG.  19. 

stances  are  found  to  crystallize  in  this  system,  the  forms  of 
which  are  generally  very  complicated.  The  crystalline  form 
of  copper  sulphate  is  shown  in  Fig.  19. 

Sixth,  or  Hexagonal,  System. — Four  axes,  three  equal 
and  in  one  plane,  making  angles  of  60°,  and  one  longer  or 
shorter  at  right  angles  to  the  plane  of  the  other  three.  The 


FIG.  20.  FIG.  21.  FIG.  22. 

hexagonal  prism,  Fig.  20,  the  regular  six-sided  pyramid, 
Fig.  21,  and  the  rhombohedron,  Fig.  22,  are  the  common 
forms  of  the  system. 

Examples  of  this  class  are :  graphite,  mercuric,  sulphide, 
sodium  nitrate,  silicic  oxide,  calcium  carbonate,  ferrous  car- 
bonate, zinc  carbonate,  etc. 


CHEMICAL  OPERATIONS. 


DEFINITIONS. 

96.  Allotropy. — When   an   element   occurs   in  two  or 
more  distinct  forms  it  is  said  to  exhibit  allotropy,  and  the 
less  common  varieties  are  said  to  be  allotropic  forms  of  that 
element. 

97.  Chemical    Separation. — The    chemist   frequently 
finds,  in  the  course  of  experiments,  that  it  is  necessary  to 


82  THEORETICAL  CHEMISTRY.  §  5 

separate  bodies  from  one  another  ;  the  processes  employed 
for  this  purpose  are  varied,  depending  on  the  nature  of  the 
substances.  As  in  many  future  operations  it  will  be  neces- 
sary to  use  one  or  more  of  them,  it  is  advisable  that  at  this 
stage  they  should  be  studied.  As  a  rule,  one  of  the  elements 
or  compounds  in  a  mixture  possesses  some  particular  prop- 
erty that  the  other  does  not  ;  thus,  one  may  be  soluble  in 
water,  and  the  other  insoluble,  which  at  once  affords  a  means 
of  separating  them.  The  principal  methods  of  separation 
employed  are: solution,  decantation,  filtration,  crystallization, 
evaporation,  distillation,  sublimation,  and  ignition. 

98.     Solution,     Decantation,    and    Evaporation. — A 

mixture  of  two  bodies,  only  one  of  which  is  soluble  in  water, 
is  separated  by  the  action  of  that  solvent. 

EXPERIMENT. — Take  some  mixture  of  sand  and  salt,  place  in  a  test 
tube,  add  water,  and  shake  up;  allow  the  sand  to  subside,  the  clear 
solution  on  the  top  is  to  be  carefully  poured  off,  without  disturbing  the 
sediment,  into  a  porcelain  basin  (evaporating  basin).  The  process  is 
known  as  decantation. 

Place  a  piece  of  wire  gauze  on  one  of  the  rings  of  the  retort  stand, 
adjusted  at  a  suitable  height  above  the  Buusen  burner ;  put  the  evapo- 
rating basin  containing  the  solution  of  salt  on  the  gauze  and  light  the 
burner,  keeping  the  flame  small ;  the  water  will  gradually  evaporate 
and  crystals  of  salt  will  form.  The  water  has  been  driven  off  by  evap- 
oration. (The  object  of  placing  the  wire  gauze  under  the  basin  is  to 
prevent  the  flame  coming  in  actual  contact  with  and  cracking  it). 

In  cases  where  both  bodies  are  insoluble  in  water,  some  substance 
has  to  be  selected  that  acts  on  the  one  and  not  the  other. 

EXPERIMENT. — Place  some  of.  a  mixture  of  sand  and  finely  powdered 
marble  in  a  test  tube,  add  some  distilled  water,  and  heat.  Take  out  a 
few  drops  of  the  water,  and  evaporate  them  on  a  piece  of  clean  platinum 
foil ;  they  entirely  disappear,  or  only  leave  the  very  slightest  stain  on 
the  foil.  This  is  the  usual  test  employed  to  ascertain  whether  a  solvent 
has  dissolved  anything  or  not.  Decant  off  the  water  and  add  dilute 
hydrochloric  acid  to  the  mixture ;  effervescence  occurs.  When  this  is 
over,  again  take  out  a  few  drops  of  the  liquid  and  evaporate  on  the  foil ; 
a  considerable  residue  will  remain.  The  best  way  to  get  a  few  drops 
out  of  a  test  tube  is  to  put  in  the  end  of  a  glass  tube ;  then  close  the 
top  with  the  finger  and  withdraw  it.  On  removing  the  finger  the  small 
portion  it  contains  runs  out. 

The  separation  in  this  experiment  depends  on  the  fact  that  marble 
is  dissolved  by  hydrochloric  acid,  while  sand  is  not  affected  by  it. 


THEORETICAL  CHEMISTRY. 


83 


99.  Filtration. — Cases  frequently  occur  in  which  decan- 
tation  is  not  advisable  or  practicable  for  the  purpose  of 
separating  a  liquid  and  a  solid ;  the  solid,  for  instance,  may 
be  so  finely  divided  that  it  floats  in  the  liquid,  giving  it  a 
muddy  appearance.  Under  such  circumstances  the  chemist 
must  resort  to  filtration. 

EXPERIMENT. — Take  a  piece  of  filter  paper  (soft  blotting  paper  will 
do,  if  you  have  not  the  especially  prepared  filter  paper  at  hand),  about 


FIG.  23. 

3|  inches  in  diameter,  and  double  it  twice ;  then  open  it  into  a  cone, 
taking  three  folds  of  paper 
on  the  one  side  and  one  on 
the  other,  as  shown  in  Fig. 
23.  Place  this  cone  of  paper 
in  a  glass  funnel,  which  it 
will  just  fit,  and  moisten  it 
with  water.  Care  must  be 
taken  that  the  point  of  the 
folded  filter  does  not  get 
broken.  Then  mix  a  solution 
of  calcium  chloride  with  some 
ammonium  carbonate;  a 
white  precipitate  will  result. 
Pour  the  whole  on  the  filter, 
holding  a  glass  rod  or  thin 
piece  of  glass  tubing  against 
the  lip  of  the  beaker  or  test 
tube  down  which  the  liquid 
runs  (see  Fig.  24) ;  the  clear 
liquid,  known  as  the  filtrate, 
passes  through,  and  is  col- 
lected in  a  bottle  or  beaker ; 


84  THEORETICAL  CHEMISTRY.  §  5 

the  precipitate,  which  consists  of  calcium  carbonate,  remains  on  the 
filter.  Pour  some  clean  water  on  it  for  the  purpose  of  washing.  After 
this  has  drained  off,  remove  the  beaker,  punch  a  hole  through  the 
bottom  of  the  filter,  wash  the  precipitate  into  an  evaporating  basin  and 
dry  it.  The  filtrate  contains  ammonium  chloride,  which  may  be  obtained 
by  evaporation. 

100.  Crystallization. — Where  two  bodies   are   mixed 
together,  both  of  which  are  soluble  in  water,  but  in  different 
degrees,  another  plan  may  be  adopted  for  their  separation. 

EXPERIMENT. — Take  a  mixture  of  about  equal  parts  of  potassium 
chlorate  and  potassium  chloride.  Place  some  in  a  test  tube,  and  add 
sufficient  water  to  dissolve  the  whole  on  boiling;  now  allow  the  solu- 
tion to  cool ;  tabular  crystals  consisting  of  the  less  soluble  potassium 
chlorate  will  separate  out.  When  cold,  filter  these  off,  wash  with  cold 
water,  and  dry  at  a  gentle  heat. 

Separation  by  crystallization  is  a  process  largely  used  for 
manufacturing  purposes  and  is  frequently  met  in  practical 
work. 

101.  Distillation. — In  the  experiment  made  by  evap- 
oration, we  have  only  dealt  with  solids  that  remain  behind, 
but  in  many  cases  it  is  the  liquid  we  wish  to  obtain.     In  the 
laboratory,  where  pure  water  is  an  essential,  it  is  obtained 


FIG.  25. 


§  5  THEORETICAL  CHEMISTRY.  85 

by  the  process  of  distillation,  in  which  the  steam  is  again 
condensed  and  collected. 

EXPERIMENT. — Take  either  an  ordinary  flask  fitted  with  cork  and 
delivery  tube,  or,  preferably,  a  flask  with  side  tube,  as  shown  at  A  B, 
in  Fig.  25.  Half  fill  it  with  brine  to  which  a  few  drops  of  coloring 
matter  (ink  will  do  nicely)  have  been  added ;  fix  it  in  the  retort  stand 
and,  as  shown,  connect  the  tube  B  by  a  cork  to  a  Liebig's  condenser 
CD.  The  condenser  consists  of  two  glass  tubes,  one  fitting  inside  the 
other;  the  steam  passes  through  the  inner  one,  and,  as  condensed, 
runs  into  the  flask  (7,  placed  as  a  receiver.  Through  the  outer  tube  a 
current  of  cold  water  is  passed,  which  rapidly  condenses  the  steam. 
The  pipe  E,  conveying  the  water,  is  attached  to  the  lower  end  of  the 
condenser;  by  pipe  Fthe  overflow  water  is  led  to  a  drain  or  into  a  pan 
or  pail  placed  conveniently.  After  some  water  has  collected  in  the 
receiver,  taste  it,  and  observe  that  it  is  perfectly  free  from  salt  and 
coloring  matter. 

A  simple,  home-made  apparatus  is  shown  in  Fig.  26.  It 
consists  of  two  flasks  A  and  C  and  a  piece  of  glass  tubing 
about  18  inches  long,  bent  at  one  end  and  fitted  through  a 


FIG.  26. 


cork  into  flask  A.  To  hasten  condensation,  a  piece  of  wet 
cloth  is  tied  round  the  glass  tubing  B  and  kept  wet  with  cold 
water  as  long  as  the  process  of  distillation  lasts. 


86  THEORETICAL  CHEMISTRY.  §  5 

102.  Sublimation. — This  is  a  process  somewhat  analo- 
gous to  distillation;  it  is,  in  fact,  distillation  of  substances 
that  condense  into  the  solid  instead  of  the  liquid  state. 

EXPERIMENT. — Heat  a  small  portion  of  a  mixture  of  sand  and  am- 
monium chloride  in  a  test  tube ;  dense  white  fumes  are  evolved,  which 
condense  as  a  white  crust  in  the  upper  part  of  the  tube,  the  sand 
remaining  behind. 

Bodies  that  distil  or  sublime  are  called  volatile ;  and  those 
that  neither  distil  nor  sublime  are  said  to  be  fixed. 

103.  Ignition. — Occasionally  the  chemist  makes  use  of 
the  combustibility  of  bodies,  and  burns  them  off  from  others 
that  are  not  combustible. 

EXPERIMENT. — Mix  some  sand  and  lampblack,  place  some  of  it  on  a 
piece  of  platinum  foil,  and  keep  at  bright  red  heat  with  the  Bunsen 
burner ;  the  lampblack  will  soon  burn  off,  leaving  the  incombustible 
sand  behind. 

It  may  here  be  mentioned  that,  when  the  directions  are 
given  to  ignite  a  body,  it  simply  means  that  it  is  to  be  heated 
intensely,  but  it  does  not  always  follow  that  it  will  burn. 

104.  Combustion. — Every  one  is  familiar  with  the  term 
combustion,  which  is  generally  considered  as  an  act  of  burn- 
ing.    We  may  now  inquire  a  little  more  closely  into  what 
chemists  understand  by  this  expression.     It  has  been  previ- 
ously mentioned  that  heat  is  always  produced  when  chemical 
combination  occurs.     If,  for  instance,  we  add  water  to  sul- 
phuric acid,  a  considerable  elevation  of  temperature  results 
from  the  combination.     This,  of  course,  is  only  one  of  the 
many  cases  that  could  serve  as  an  example,    there  being 
many  others  in  which  the  heat  is  far  more  intense. 

Whenever  the  heat  caused  by  chemical  union  is  sufficiently 
intense  to  raise  the  products  to  a  temperature  at  which  they 
emit  light,  the  act  of  union  is  termed  combustion. 

The  meaning  of  the  term,  however,  has  been  still  further 
widened;  thus,  certain  processes  of  decay,  and  also  the 
chemical  changes  that  produce  animal  heat,  are  cases  of  com- 
bination with  oxygen.  These  chemical  actions  are  very 
often  termed  instances  of  sloiv  combustion. 


§  5  THEORETICAL  CHEMISTRY.  87 

105.  Destructive  Distillation. — Destructive  distilla- 
tion may  be  defined  as  the  resolution  of  a  complex  substance 
into  simpler  vapors,  gases,  and  liquid  products  under  the 
influence  of  heat,  out  of  contact  with  the  air.     The  operation 
of  destructive  distillation  will  be  fully  described  in  the  intro- 
duction to  Organic  Chemistry. 

106.  Explosion. — When  two  or  more  substances,  one 
of  which  is  inflammable  and  the  others  supporters  of  com- 
bustion, are  mixed  and  ignited,  combination  between  these 
bodies  takes  place  with  extreme  rapidity.     The  heat  evolved 
increases  the  pressure  of  the  gases  produced  as  a  result  of 
the  chemical  action,  and  these  give  the  surrounding  air  such 
a  severe  shock  as  to  cause  a  loud  detonation. 

Extremely  rapid  combustion,  attended  with  more  or  less 
noise,  is  termed  explosion;  and  the  mixture  that  thus  ex- 
plodes is  called  an  explosive  body. 

107.  Inflammable  Bodies  and  Supporters  of  Com- 
bustion.— Chemistry  has  divided  all  bodies  that  take  part  in 
the  act  of  combustion  into  inflammable  bodies  and  supporters 
of  combustion.     Thus,  when  hydrogen  burns  in  oxygen,  the 
former  is  said  to  be  inflammable,  while  the  latter  is  said  to 
support  the  combustion. 

These  terms,  however,  are  simply  relative.  We  chance  to 
live  in  an  atmosphere  of  which  oxygen  is  the  most  active 
constituent;  when  a  jet  of  hydrogen  is  burned,  what  takes 
place  is  that  at  the  opening  of  the  jet  the  hydrogen  and  the 
oxygen  of  the  air  come  in  contact  and  combination  occurs. 
If  the  atmosphere,  by  chance,  should  consist  of  hydrogen, 
then  that  element  would  be  classified  as  a  supporter  of  com- 
bustion, and  oxygen  and  other  similar  bodies  would  be 
inflammable  substances. 

108.  Nascent  Condition  of  Elements.— The  student 
already  knows  that  the  term  atom  is  applied  to  the  smallest 
particles  of  an  element,  and  that  these  elements  unite  to 
form  elemental  molecules.     Now,  we  may  assume  that  some 
portion  of  the  energy  that  tends  to  force  these  atoms  to  unite 


88  THEORETICAL  CHEMISTRY.  §  5 

with  other  atoms  is  consumed  when  several  atoms  of  the 
same  kind  unite  to  form  an  elemental  molecule ;  in  fact,  it 
has  been  found  that  elements  in  the  atomic  state  possess  a 
greater  combining  energy  and  are  more  active  than  the  same 
elements  in  the  form  of  elemental  molecules. 

This  assumption  may  be  proved  to  be  correct  by  the  follow- 
ing experiment: 

EXPERIMENT. — Dissolve  some  ferric  chloride,  which  possesses  a 
natural  red  color,  and  allow  hydrogen  to  bubble  up  through  the  solu- 
tion ;  you  will  see  that  there  is  no  alteration  in  the  appearance  of  the 
ferric  chloride.  Now,  add  some  sulphuric  acid  and  a  few  pieces  of 
zinc,  and  you  will  notice  that  the  color  of  the  ferric  chloride  very 
quickly  disappears. 

Why  does  this  happen?  The  explanation  is  very  simple: 
in  the  second  case — that  is,  when  sulphuric  acid  and  zinc  were 
added  to  the  ferric-chloride  solution — hydrogen  atoms  were 
evolved  within  the  solution,  and  the  liberated  atoms,  instead 
of  combining  to  form  hydrogen  molecules,  acte,d  upon  the 
ferric  chloride  with  considerable  energy.  The  following 
equations  show  the  chemical  changes  that  occur: 

Zn     +     HSO. 


(2) 


The  reducing  power  of  nascent  hydrogen  may  be  more 
strikingly  illustrated  by  substituting  potassium  permanga- 
nate and  hydrochloric  acid  for  ferric  chloride  and  sulphuric 
acid,  in  the  above  experiment.  The  nascent  hydrogen  in 
this  case  will  reduce  the  potassium  permanganate,  and  thus 
destroy  the  strong  color  it  imparts  to  a  solution. 

An  essential  feature  of  this  type  of  chemical  action  is  that 
it  is  caused  by  an  element  at  the  moment  it  is  liberated  from 
a  compound.  Consequently,  at  the  moment  of  liberation 
from  a  compound,  an  element  is  said  to  be  nascent.  (Nascent 
means  "being  born. ")  Most  elements  possess  more  active 
chemical  properties  when  in  the  nascent  state. 


**ffl'         1         -'''24 

zinc           sulphuric 

zinc              nascent 
sulphate        hydrogen 

2ff    +    Fefl.    = 

nascent           ferric 
hydrogen       chloride 

ZFeCt,    +     1HCI 

ferrous        hydrochloric 
chloride               acid 

§  5  THEORETICAL  CHEMISTRY.  89 

When  elements  are  in  the  molecular  state,  the  force  by 
which  the  atoms  are  held  together  must  be  overcome  before 
they  can  enter  into  fresh  combinations.  When  they  are  in 
the  nascent  condition  no  such  force  has  to  be  overcome,  and 
so  the  atoms  are  able  to  enter  into  combination  with  much 
more  energy,  none  of  their  original  energy  having  previously 
been  absorbed. 

1O9.  Oxidation. — The  compounds  of  oxygen  with  other 
elements  are  called  oxides,  and  the  act  of  combination  is 
termed  oxidation.  This  process  is  generally  a  slow  one  and 
its  effects  are  not  immediately  perceptible.  Some  familiar 
examples  are  the  tarnishing  or  rusting  of  metals  exposed  to 
the  air,  the  gradual  decay  of  wood,  the  drying  of  oils  in 
paint,  the  formation  of  vinegar  from  alcohol,  the  respiration 
of  animals,  and  combustion. 

In  all  these  processes  heat  is  generated;  but  it  is  not 
usually  noticed  unless  it  is  sufficient  to  render  the  matter 
luminous,  which,  as  we  have  already  seen,  is  the  case  only 
with  combustion.  The  term  oxidation  is  frequently  used  in 
another  sense.  We  have  seen  that  many  of  the  elements 
have  variable  valence,  and  certain  metals,  on  account  of  this, 
form  two  series  of  salts.  Thus  iron  forms  ferrous  and  ferric 
salts,  in  which  its  valence  is  apparently  2  and  3,  respectively. 
A  change  from  a  lower  to  a  higher  valence  is  called  oxida- 
tion, although  oxygen  may  not  be  present.  Thus  we  speak 
of  oxidizing  ferrous  chloride  to  ferric  chloride,  or  of  chan- 
ging from  a  lower  to  a  higher  state  of  oxidation. 

HO.  Oxidizing  Agents. — An  oxidizing  agent  may  be 
defined  as  a  substance  that  causes  an  element  or  compound 
to  combine  with  oxygen  or  with  other  elements  of  similar 
character  or  increases  its  valence.  Oxygen  and,  to  a  greater 
degree,  ozone  are  the  most  striking  examples  of  oxidizing 
agents.  If,  for  instance,  metallic  copper  is  heated  in  a  cur- 
rent of  oxygen,  the  following  reaction  takes  place: 

2Cu     +     <92     =     2CuO 
copper         oxygen     copper  oxide 


90  THEORETICAL  CHEMISTRY.  §  5 

The  following"  substances  are  some  of  the  most  frequently 
occurring  oxidizing  agents:  oxygen,  ozone,  chlorine,  bromine, 
potassium  chlorate,  nitric  acid,  and  hydrogen  peroxide.  (The 
special  properties,  etc.  of  these  substances  will  be  fully 
treated  in  Inorganic  Chemistry.) 

111.  Reaction  and  Reagent.  —  The  term  reaction  is 
applied  to  any  chemical  change  that  may  be  produced,  and 
the  substance  applied  to  produce  the  change  is  termed  a 
reagent. 


Reducing  Agents.  —  The  native  compounds  or 
ores,  from  which  metals  are  derived,  are  in  many  cases 
oxides  or  sulphides  of  such  metals.  The  process  used  to 
obtain  these  metals  in  a  free  or  uncombined  state  is  known 
as  the  reduction  of  the  ore  to  its  metallic  state.  From  this 
we  define  a  reducing  agent  as  a  substance  that  removes  oxy- 
gen, or  elements  similar  to  it,  from  compounds,  or  decreases 
the  valence  of  a  substance,  changing  it  from  a  higher  to  a 
lower  state  of  oxidation. 

Owing  to  the  activity  with  which  hydrogen  combines  with 
oxygen,  it  is  one  of  the  most  prominent  and  powerful  redu- 
cing agents  with  which  we  are  acquainted.  Thus,  if  hydro- 
gen is  passed  over  red-hot  copper  oxide  or  iron  oxide,  the 
metal  is  obtained  in  the  free  state,  according  to  the  following 
equations  : 

(1)  CuO    +     H9     =     Cu     +     H9O 

opper      hydrogen      copper          water 

(2)  Ft,0,    +    3H,    =    fe,    +    Sff,O 

iro?  hydrogen         iron  water 

It  may  here  be  mentioned,  incidentally,  that  in  this  case 
hydrogen,  a  g-as  that  may  be  produced  by  the  reducing 
action  of  iron  on  steam,  is  passed  over  copper  oxide,  a  prod- 
uct of  the  oxidizing  influence  of  the  oxygen  of  the  air  on 
copper. 

The  reaction  of  hydrogen  on  copper  oxide  is  of  great  im- 
portance, as,  in  the  analysis  of  many  compounds  containing 


§  5  THEORETICAL  CHEMISTRY.  91 

hydrogen,  the  hydrogen  is  thus  converted  into  water  and 
weighed  as  such.  In  addition  to  these  instances,  in  which  a 
reducing  agent  is  employed  in  order  to  obtain  metals  in  their 
free  state,  there  are  other  examples  of  reduction  in  which 
bodies  are  simply  reduced  to  a  lower  state  of  oxidation. 
Further,  the  same  reducing  agents  are  capable  of  removing, 
in  whole  or  in  part,  as  the  case  may  be,  chlorine  and  other 
similar  elements  from  compounds.  Among  the  most  active 
reducing  agents,  in  addition  to  hydrogen,  are :  carbon,  carbon 
monoxide,  stannous  chloride,  and  sulphurous  acid. 


ACIDIMETRY   AKD    ALKALIMETRY. 

113.  Among  the  various  operations  performed  in  analyt- 
ical work,  one  of  the  most  important  series  is  that  dealing 
with  the  estimation  of  the  quantities  of  acid  or  alkali  con- 
tained in  various  substances.     The  general  practice  is  to  add 
to  a  measured  quantity  of  the  substance  an  acid  or  alkali 
solution  of  known  strength  until  the  point  of  neutrality  is 
reached.      These  test  solutions  are  added  from  graduated 
vessels  known  as  burettes,    and  the  neutral  point  is  deter- 
mined by  the  use   of  some  indicating  substance.     The  best 
known  of  these  are :  litmus,  methyl-orange,  and  pJienol-pJitha- 
lein. 

114.  Litmus. — It  has  already  been  mentioned  that  lit- 
mus is  a  vegetable  salt,  which  is  turned  red  by  acids  and  has 
its  original  blue  color  restored  by  alkalies. 

115.  Methyl- Orange. — Methyl-orange    is    used   in    a 
solution  of  alcohol,  and  in  alkaline  solutions  possesses  a  yel- 
low color,  which,  however,  changes  to  reddish  pink  with  the 
slightest  excess  of  acid.      Since  methyl-orange  is  not  in  the 
least  affected  by  carbonic  acid,  it  is  of  considerable  value  in 
certain  estimations. 

116.  Phenol-Phthalein.  —  Phenol-phthalein    is    more 
sensitive    than   litmus,   and   in   an   acid   solution   does  not 


92  THEORETICAL  CHEMISTRY.  §  5 

produce  a  color,  but  with   the  slightest  excess  of  alkali  an 
intense  magenta  red  is  produced. 


117.  Standard  Solutions.  —  A  series  of  standard  solu- 
tions is  used  in  acidimetry  and  alkalimetry,  the  solutions  being 
of  a  strength  known  in  the  laboratory  as  normal  solutions. 
The  normal  acid  solutions  are  so  prepared  that  at  16°  1  liter 
of  the  acid  contains  1  gram  of  replaceable  hydrogen. 

Accordingly,  the  normal  sulphuric-acid  solution  would 
contain  : 


(2  +  32  -f-  64)  -h  2  =  49  grams  per  liter,  or  .  049  gram  per 
cubic  centimeter;  and  the  normal  hydrochloric-acid  solution 
would  contain: 


Cl 

1  +35.37  =  36.37  grams  per  liter,  or  .036  gram  per  c.  c. 

Each  liter  of  the  normal  alkali  solutions  must  contain  the 
chemical  equivalent,  weighed  in  grams,  of  the  hydrogen- 
replacing  metal  or  group  of  elements. 

Therefore,  the  normal  sodium-  hydrate  solution  must 
contain  : 

Na  +  H+O 

23  +  1  -f-  16  =  40  grams  per  liter,  or  .040  gram  per  c.  c. 
(Sodium  and  hydrogen  are  equivalent  to  each  other.  ) 
The  normal  barium-hydrate  solution  BaH^O^  must  contain: 


(137+  2  -f-32)-r-2  =  85.5  grams  per  liter  =  .085  gram  per 
cubic  centimeter. 

(The  equivalence  of  barium  is  twice  that  of  hydrogen.) 

118.  Equal  quantities  of  any  normal  acid  and  alkali  solu- 
tion neutralize  each  other  ;  and  the  quantity  of  any  solution 
of  a  single  acid  or  alkali  that  is  neutralized  by  1  cubic  centi- 
meter of  a  normal  solution  contains  .  001  gram  of  the  hydrogen 
equivalent  of  its  active  constituent. 


THEORETICAL  CHEMISTRY. 


93 


The  following  description  of  a  typical  experiment  will  fully 
explain  the  operations  performed : 

EXPERIMENT. — The  necessary  equipment  consists  of  some  standard 
normal  solution  of  sulphuric  acid  and  sodium  hydrate,  a  double  burette 
stand  and  two  burettes,  each  of  50  cubic  centimeters  capacity.  A 
burette  consists  of  a  glass  tube  open  at  the  top  and  drawn  to  a  point 
at  the  bottom,  over  which  is  drawn  a  piece  of  rubber  tube  with  a  small 
glass  dropping  tube  at  the  end  of  it.  A  spring  clip  (pinch  cock) 
permits  the  bottom  of  the  burette  to  be  opened  or  closed  at  will.  The 
instrument  is  graduated,  having  the  zero  at  the  top  and  the  scale  of 
figures  descending.  The  burette  stand,  with  the  two  burettes,  is  shown 
in  Fig.  27,  and  the  end  of  a  burette  in  Fig.  28. 

After  carefully  cleaning  the 
two  burettes  and  allowing  them 
to  drain,  one  is  rinsed  and  sub- 
sequently filled,  above  the  zero 
mark,  with  normal  acid  solu- 
tion, and  the  other  one  is  rinsed 
and  then  filled  above  the  zero 
mark  with  normal  sodium- 
hydrate  solution.  Then  we 
press  the  buttons  of  the  spring 
clip  in  each  case  until  the  liquid 
runs  down  to  the  zero  mark. 
Next,  we  pour  about  50  cubic 
centimeters  of  distilled  water 
into  a  clean  porcelain  basin 
about  4  inches  diameter,  and 
run  into  it  from  the  burette  10 
cubic  centimeters  of  normal 
acid  solution.  Then  we  add 
two  or  three  drops  of  phenol- 
phthalein  solution  and  notice 
that  there  is  no  color  produced. 
We  read  off  the  height  of  the 
soda  (sodium-hydrate)  solution 
and  make  a  note  of  it.  Then 
without  consulting  the  gradua- 
tions, we  run  in  the  soda  solu- 
tion, drop  by  drop,  until  one 
drop  imparts  a  red  .color  to  the 
solution.  If  we  now  read  off 
the  height  of  the  soda  solution 
in  the  burette,  we  should  find  that  exactly  10  cubic  centimeters  have 
been  used  to  neutralize  the  acid. 


FIG. 


94 


THEORETICAL  CHEMISTRY. 


Then  we  repeat  the  operation,  but  ask  some  one  to  run  a  quantity 
of  the  acid,  to  us  unknown,  into  the  basin.  We  test  with  soda  solution, 
reading  off  and  carefully  noting  the  height  of  the  burette  before 
the  solution  is  added,  and  again  as  soon  as  the  red  color  is  produced ; 
and  then  we  compare  the  quantity  of  the  acid  and  soda  solutions, 
and  see  whether  they  agree. 

We  further  make  a  similar  determination  of 
alkali  by  working  with  an  unknown  quantity 
and  adding  acid  until  the  color  of  the  phenol- 
phthalein  disappears. 

Next  we  determine  the  percentage  of  sulphuric 
acid  in  a  dilute  solution  of  unknown  strength. 
We  take  a  small  beaker,  carefully  wipe  it  dry, 
and  weigh  it.  We  pour  into  it  about  10  cubic 
centimeters  of  the  dilute  acid,  and  weigh  it 
again  carefully.  Then  we  add  some  distilled 
water  and  two  or  three  drops  of  phenol-phthalein, 
and  then  titrate  with  normal  sodium-hydrate 
solution,  and  calculate  the  percentage  of  acid 
present.  (The  verb  titrate  means  to  measure 
by  means  of  solutions  of  known  strength,  such 
as  are  here  employed.) 

In  the  next  place,  we  determine  the  percentage 
of  sodium  hydrate  in  a  similar  manner  in  a 
dilute  solution  of  unknown  strength,  titrating 
with  normal  sulphuric  acid. 

Next  we  determine  the  percentage  of  sodium 
carbonate  in  a  sample  of  common  washing  soda. 
We  weigh  off  exactly,  say  2  grams  of  the  soda 
and  dissolve  it  in  50  cubic  centimeters  of  dis- 
tilled water  in  a  small  beaker;  we  add  two  or 
three  drops  of  methyl-orange,  and  titrate  with 
normal  acid  until  the  yellow  tint  changes  to 

red.     With  each  addition  of  acid  there  is  an  effervescence  of  carbon 
dioxide,  which,  however,  does  not  interfere  with  the  estimation. 

These  calculations  are  best  illustrated  by  means  of  analo- 
gous calculations.  The  following  are  results  actually 
obtained  for  this  purpose : 


DETERMINATION  OF  SULPHURIC  ACID. 

Weight  of  beaker  and  acid 41 . 92  grams. 

Weight  of  dry  beaker 30.23  grams. 

Weight  of  acid 11 .69  grams. 


§  5  THEORETICAL  CHEMISTRY.  95 

Final  reading"  of  burette 25 . 9  c.  c. 

First  reading  of  burette 8 . 4  c.  c. 

Normal  solution  used 17.5  c.  c. 

Normal  acid  and  soda  are  equivalent  to  each  other,  and 
therefore  11.69  grams  of  the  acid  solution  to  be  tested  con- 
tains sulphuric  acid  equal  to  the  quantity  in  17.5  cubic 
centimeters  of  standard  acid ;  that  is, 

17.5X.049  =  .8575  gram. 

.8575-7-11.69  =  .0733;  therefore  the  solution  contains  7.33 
per  cent,  of  sulphuric  acid. 


DETERMINATION  OF  SODIUM  CARBONATE. 

Weight  of  washing-soda 2.03  grams. 


Final  reading  of  burette  ............   43  .  3  c.  c. 

First  reading  of  burette  ............   31  .  1  c.  c. 

Acid  used  for  neutralization.  .    12.2  c.  c. 

It  is  first  necessary  to  decide  what  quantity  of  sodium  car- 
bonate is  equivalent  to  the  normal  solution  of  sulphuric 
acid.  The  following  are  the  formulas  and  molecular  weights 
of  anhydrous  and  crystallized  carbonates,  respectively  : 


46  +  12  +  48  =  106,  molecular  weight  of  anhydrous  sodium 
carbonates. 


46  +  12  +  48  +  10(24-16)  =  286,  molecular  weight  of  crys- 
tallized sodium  carbonate. 

As  the  molecule  contains  twice  the  sodium  equivalent,  it 
is  evident  that  half  of  these  quantities  —  namely,  53  grams 
per  liter  of  the  anhydrous,  and  143  grams  for  the  crystallized 
carbonates  —  are  the  strengths  for  normal  solutions  equivalent 
to  those  of  other  acids  and  alkalies.  From  this  it  follows 
that  the  values  for  cubic  centimeters  are,  respectively,  .053 


96  THEORETICAL  CHEMISTRY.  §  5 

and  .  143  gram.  The  acid  taken  for  neutralization  is,  there- 
fore, equivalent  to: 

12.2  X. 053  =  .447  gram  of  N 
12.2  X. 143  =  1. 745  grams  of 

From  these  the  percentages  are  readily  obtained: 

As  2.03  :  100  =  .447  :  22.02  per  cent,  of  NafO%  in  sample. 
As  2.03  :  100  =  1.745  :  85.96  per  cent,  of  NafO^HJ)  in 
sample. 

Further,  100  —  85.96  =  14.04  per  cent  of  water  and  other 
impurities. 

(A  purposely  contaminated  sample  had  been  used.) 


LABORATORY  DIRECTIONS  AND  PRACTICAL 
HINTS. 

119.  Several  of  the  experiments  described  in  this  paper, 
although  illustrating  only  simple  facts,  require  some  skill 
and  experience  in  chemical  manipulations  for  their  success- 
ful performance,  and  the  student  will  probably  be  content 
with  carefully  studying  them  and  grasping  the  lesson  that 
they  convey.  It  should,  however,  be  mentioned  that  exper- 
imenting is  an  essential  to  the  study  of  chemical  science, 
and  we  advise  the  student  to  try  to  perform  as  many  of  the 
experiments  given  in  the  following  Instruction  Papers  as 
possible. 

The  student  will,  after  a  few  unsuccessful  trials,  soon  gain 
some  experience;  he  should,  however,  always  remember 
that  "patience  and  carefulness "  are  indispensable  to  the 
make-up  of  a  good  chemist. 

If  you  are  not  successful  with  an  experiment,  try  again 
and  again  until  you  succeed. 

To  facilitate  the  study  of  chemistry  for  those  that  have 
not  access  to  a  laboratory,  THE  INTERNATIONAL  CORRESPOND- 
ENCE SCHOOLS  furnish  a  set  of  apparatus,  and  a  large  num- 
ber of 'chemicals  and  reagents  at  a  very  moderate  price. 


THEORETICAL  CHEMISTRY. 


97 


12O. 

ments: 


List  of  apparatus  for  the  performance  of  experi- 


1  Iron  Triangle. 

1  Pair  Forceps,  5  in. 

1  Iron  Triangle  with  Clay  Tubes. 

1  Shallow  Iron  Sand  Bath,  4  in. 

1  Brass  Blowpipe  with  Bulb,  8  in. 

1  Piece  of  Charcoal. 

1  Piece  Platinum  Wire. 

2  Pieces  Wire  Gauze,  4  in. 

1  doz.  Corks,  Assorted. 

2  feet  Rubber  Tubing,  /,  in. 

1  Deflagrating  Spoon. 

2  Sheets   Red   and   Blue    Litmus 
Paper. 

Platinum  Foil,  1  in.  X  2  in.  X  WW 

in.  thick. 

1  Woulff  Bottle,  3  necks,  8  oz. 
1  French-Clay  Crucible,  No.  4. 
1  Wrought-Nickel  Crucible,  20  c.  c. 
1  Measuring  Cylinder,  25  c.  c. 
1  Graduated  Test  Tube,  5  c.  c. 
1  Stirring  Rod. 

1  Piece  Blue  Glass. 

2  U  Tubes,  6  in. 

1  Pkg.  J.  H.  Munktell's  No.  0  Fil- 
ters, 9  cm. 
1  Pipette,  1  c.  c. 
1  Pipette,  5  c.  c. 
1  Thermometer,  graduated  to200°C. 


The  room  chosen  for  a  laboratory  should  be  of  suf- 
ficient height,  well  lighted  and  ventilated.  It  should  con- 
tain a  working  bench  or  long  table  of  hard  and  durable  wood, 
some  shelves  for  bottles  and  apparatus,  gas  for  heating  and 
lighting,  and  should,  if  possible,  be  close  to  a  sink  and  water 
supply. 

The  student  should,  in  the  first  place,  make  himself  famil- 
iar with  the  names  and  uses  of  his  apparatus.  The  bottles 
and  jars  containing  chemicals,  all  properly  labeled,  should  be 
so  placed  that  they  are  easily  accessible.  They  should  never 
be  removed  from  or  allowed  to  stand  around  on  the  working 


1  Bunsen  Burner,  with  3-foot  Tu- 
bing. 

1  Nest  of  3  Beakers,  Nos.  1-3. 
3  Flasks,  4,  8,  and  16  oz. 

2  Wide-Mouth  Bottles,  \  gal. 
\  Ib.  Glass  Tubing. 

Thistle  Funnel,  12  in. 

Glass  Funnel,  2  in. 

Tubulated  Retort,  4  oz. 

Receiver,  Plain,  4  oz. 

Test  Tubes,  5  in.  X  f  in. 

Stand. 

Safety  Tube,  with  Thistle  Top. 

Porcelain  Dishes,  3i  in.  X  4|  in. 

Mortar  and  Pestle,  3£  in. 

Square  Glass  Plate,  4  in. 

Mohr's  Pinch  Cock. 

Pan,  Agate  Iron,  8  in. 

Beehive  Shelf. 

Iron  Retort  Stand,  2  rings. 
1  Clamp,  Small   Universal  Move- 
ment. 

1  Iron  Tripod. 
1  Balance,   sensitive  to  ^  gr.  (1 

mg.). 

1  Set  Weights,  50  g.  to  1  mg. 

2  Burettes,    50  c.    c.,  with    Glass 
Stop-Cock. 


98  THEORETICAL  CHEMISTRY,  §  5 

bench;  the  quantity  required  should  be  taken  in  a  flask  or 
test  tube,  and  the  bottle  should  be  immediately  restored  to 
its  proper  place.  A  strict  adherence  to  this  rule  will  prevent 
the  loss  of  time  that  otherwise  would  be  spent  in  hunting 
all  over  the  place  for  the  desired  substance,  and  will  save 
confusion. 

In  getting  the  chemicals  for  an  experiment,  be  sure  that 
the  right  ones  are  taken ;  see  that  concentrated  acids  are  not 
employed  where  dilute  acid  should  be  used,  and  never  use 
C.  P.  (chemically  pure)  substances  when  the  commercial 
ones  will  answer.  Do  not  allow  stoppers  or  corks  to  lie 
about,  or  get  mixed  up.  Take  no  more  of  the  substance 
than  you  require,  but  if  you  happen  to  have  an  excess,  do 
not  return  it  to  the  bottle.  It  is  better  to  pour  it  away,  as 
the  greatest  confusion  has  been  caused  by  chemicals  being 
replaced  in  the  wrong  bottles. 

When  directions  are  given  that  substances  should  be 
mixed,  solid  bodies  are  first  to  be  powdered  in  a  mortar, 
and  then  stirred  together  until  the  mixture  is  as  uniform  as 
possible. 

Heat  is  applied  by  means  of  the  Bunsen  burner,  and  in 
heating  test  tubes  the  heat  should  be  applied  gently  at  first, 
with  a  constant  rotary  motion  of  the  tube,  or,  if  the  tube  is 
fixed,  of  the  burner. 

If  heating  a  liquid,  never  let  the  flame  play  on  the 
part  of  the  tube  above  the  surface  of  the  liquid.  When  a 
solid  is  heated,  if  any  moisture  happens  to  be  present, 
it  condenses  in  the  upper  and  cooler  parts  of  the  tube; 
when  there  is  the  slightest  sign  of  this,  hold  the  tube 
almost  horizontal,  but  with  the  hot  part  rather  higher; 
this  prevents  the  condensed  water  from  running  back  on 
the  hot  glass. 

A  diary  or  note  book  should  be  conscientiously  kept,  and 
every  experiment,  the  changes  noticed,  and  the  results 
should  be  carefully  entered.  This  applies  with  special  force 
to  weighing. 

After  you  have  finished  your  laboratory  work,  carefully 
wash  all  the  apparatus  used,  and  wipe  with  a  dry  cloth  any 


THEORETICAL  CHEMISTRY. 


99 


water  from  the  retort  staixd  or  other  apparatus  that  might 
rust. 

Never  use  a  piece  of  apparatus  until  you  have  carefully 
dusted  it  and  have  convinced  yourself  that  it  is  perfectly 
clean. 


122.  The  Bunsen Burner. — The  Bunsen  burner,  shown 
in  Fig.  29,  is  named  after  its  inventor.  It  consists  of  a  gas 
tube  a,  which  projects 
part  way  into  a  large  tube 
b,  called  the  mixing  tube. 
Air  is  admitted  through 
holes  c,  which  are  closed 
or  regulated  by  means  of. 
the  collar  or  slide  d.  The 
gas  issuing  from  a  and 
the  streams  of  air  from 
the  holes  c  mingle  in  the 
upper  part  of  the  tube  b 
and  form  a  mixture  that 
will  burn  over  the  mouth 
of  the  tube.  Before  light- 
ing the  burner,  see  that 
the  holes  c  at  the  bottom 
are  open.  The  flame 
should  be  non-luminous; 
if  smoky,  it  is  a  sign  that 
the  gas  is  burning  at  the 
lower  end  of  the  tube  of 


the  burner,   and  it  must 

be     extinguished     and 

relighted.     This  catching  fire  at  the  bottom  is  the  result  of 

the  entrance  of  an  excess  of  air;  the  remedy  is  to  turn  on 

more  gas,  or  partly  close  the  air  holes  by  adjusting  the  slide 

so  as  to  restrict  the  air  supply. 


FIG.  29. 


123.     Glass. — Glass  is  indispensable  to  the  chemist  on. 
account  of  its  transparency,  the  ease  with  which  it  can  be 


100  THEORETICAL  CHEMISTRY.  §  5 

worked  in  any  shape  when  hot  and  plastic,  and  its  not  being 
attacked  by  most  chemicals.  Its  disadvantage  is  its  brittle- 
ness.  It  is  particularly  liable  to  break  on  heating;  being  a 
bad  conductor  of  heat,  one  part  of  the  glass  expands  on 
being  heated  before  the  neighboring  parts  become  warm. 

When  boiling  water  or  any  other  liquid  in  a  glass  vessel, 
always  use  a  tripod  and  put  a  piece  of  wire  gauze  between 
the  vessel  and  the  flame.  The  gauze  distributes  the  heat 
over  the  bottom  of  the  vessel,  and  the  liquid  can  be  heated 
until  it  boils  without  accident;  i.  e. ,  without  cracking  the 
glass.  Or,  place  some  fine  sand  in  an  iron  tray;  put  the 
tray  upon  the  tripod  stand  and  place  the  glass  vessel  upon 
the  sand.  The  heat  is  distributed  "by  the  sand,  and  the  glass 
vessel  can  be  heated  safely.  The  tray  containing  the  sand 
is  known  in  the  laboratory  as  a  sand  bath. 

124.  To  Cut  Glass  Tubes. — It  will  generally  be  pos- 
sible for  the  student  to  fit  up  the  apparatus  required  for  his 
experiments,  and  he  will  frequently  have  to  cut  a  piece  of 
glass  tubing  to  a  desired  length.  This  is  done  very  easily. 
Take  a  sharp  file  and  make  a  nick  in  the  glass  tube ;  then 


FIG.  30. 


hold  the  tube  in  both  hands ;  the  two  forefingers  should  be 
together  immediately  under  the  nick,  the  two  thumbs  above. 
Turn  the  wrists  outwards  and  downwards,  round  the 


THEORETICAL  CHEMISTRY. 


101 


forefingers  as  axes,  at  the  same  time  pulling  the  hands  apart 
(see  Fig.  30). 

125.  To  Bend  Glass  Tubing. — Hold  a  piece  of  glass 
tubing  lengthwise  along  a  flat  gas  flame  (fishtail  or  batswing, 
not  the  Bunsen  flame),  as  shown  in  Fig.  31.  Keep  it  moving 
slowly  round  and  round  and  also  to  and  fro.  As  soon  as  it 
feels  slightly  soft,  hold  it  by  one  hand  and  cease  rotating  it. 
The  free  end  will  gradually  fall  (Fig  32).  As  soon  as  it  is 


FIG.  31. 


FIG.  32. 


bent  to  the  desired  angle,  remove  it  from  the  flame.  This 
method  is  most  frequently  used,  but  as  it  is  likely  to  cause 
the  glass  to  bunch  at  the  bend  and  partially  close  the  tube, 
it  is  better  to  hold  the  tube  by  both  ends,  and  bend  them 
upward  as  soon  as  the  glass  softens.  A  smooth  uniform 
bend  is  thus  obtained. 

126.  Mistakes  to  Avoid  in  Bending  Glass.— 1.   Hold 
a  piece  of  glass  tubing  across  the  Bunsen-burner  flame.      It 
becomes  hot  in  two  places.     When  soft  bend  it.     The  glass 
will  be  found  creased  at  the  bend.     2.   Hold  a  piece  of  glass 
tubing  across  an  ordinary  flat  gas  flame.     The  tube  is  heated 
this  time  in  one  place.      Bend  it.     The  tube  is  flattened  on 
the  outside  of  the  bend  and  thickened  on  the  inside. 

127.  To  Bound  the  Ends  of  Glass  Tubing. — The  sharp 
edges  left  at  the  ends  of  the  glass  tube  must  always  be  care- 
fully  rounded,    otherwise   the   tubes   will  not   slide   easily 


102  THEORETICAL  CHEMISTRY.  §  5 

through  corks  and  are  apt  to  cut  any  rubber  tubes  that  are 
slipped  over  them.  The  ends  may  be  rounded  by  (a)  grind- 
ing the  sharp  edge  on.a  file  or  on  a  stone,  or  (b)  by  holding 
the  end  in  a  flame  until  the  glass  just  melts  at  its  edges. 

128.  To  Draw  Out  Glass  Tubing.— 1.  Holding  a  piece 
of  glass  tube  in  both  hands,  heat  it  at  one  place  in  the  Bunsen 
flame.  When  soft  remove  it  from  the  flame  and  pull.  The 
tube  is  drawn  a  little  as  at  A,  Fig.  33.  At  B  it  is  pulled  out 


PIG.  33.  FIG.  34. 

much  finer,  because  it  happens  to  be  hotter,  and  it  has 
broken  there  because  the  pull  was  too  hard.  The  lump  C 
was  in  the  cooler  portion  of  the  flame,  and  never  became 
properly  heated. 

2.  Heat  another  tube  in  a  flat  flame,  holding  it  lengthwise 
and  turning  it  slowly  round  and  round.     Pull  the  two  ends 
apart ;  there  is  no  lump,  because  the  tube  was  heated  evenly, 
but  the  finely  drawn-out  tube  has  softened  at  once,  and  bends 
down,  as  shown  in  Fig.  34. 

3.  Heat  another  tube  as  in  Experiment  2  until  the  tube 
feels  limp  in  the  fingers.     Remove  it  from  the  flame.     Draw 
slowly  at  first,  then  more  quickly,  gently  rotating  each  end, 
until  the  hands  are  wide  apart.     Hold  rigidly  for  a  few  sec- 
onds   until     the     glass 
stiffens.    The  glass  tube 

FlG-  M-  is  now  drawn  out,  per- 

haps a  yard  long,  and  has  the  appearance  as  shown  in 
Fig.  35.  Cut  the  tube  lightly  with  a  sharp  file  at  A  and  B, 
and  notice  how  elastic  the  section  A  B  is ;  it  can  be  bent 
nearly  to  a  circle. 

129.  Rubber  tubing  sometimes  becomes  stiff  when  not 
in  use.  Warm  gently  and  pull  it  about  before  use  ;  it  soon 
softens.  If  a  rubber  tube  is  too  wide  for  a  glass  tube  it  can 
be  made  to  fit  by  doubling  it  back  on  itself. 


§  5  THEORETICAL  CHEMISTRY.  103 

1 3O.  Pipettes. — Liquids  are  measured  by  the  flasks  and 
measuring  jars  that  contain  them  ;  it  is  extremely  useful, 
however,  to  have  vessels  that  will  measure  out, 

or  deliver  accurately  a  definite  small  volume  of 
liquid. 

A  pipette  is  a  glass  tube  open  at  both  ends, 
upon  which  a  cylindrical  bulb  has  been  blown 
(see  Fig.  36).  There  is  a  circular  mark  upon  the 
stem.  When  filled  with  liquid  up  to  the  mark, 
the  pipette  will  deliver  exactly  that  volume  which 
is  marked  upon  it. 

How  to  Use  a  Pipette. — Place  the  pointed 
end  of  the  pipette  in  water.  Suck  the  water  up 
until  it  is  above  the  mark.  Quickly  close  the  end 
with  the  moist  forefinger.  Pressing  gently,  rotate 
the  pipette  so  that  the  water  very  slowly  flows 
out  until  the  water  level  reaches  the  mark.  Then 
press  lightly  ;  the  flow  ceases.  Transfer  the 
pipette  to  a  beaker,  remove  the  finger,  and  let  the 
liquid  flow  out.  As  soon  as  the  pipette  appears 
to  be  empty,  let  it  drain  for  a  few  seconds  with 
the  tip  just  below  the  surface  of  the  water  in  the  beaker. 
The  drop  of  water  that  is  still  in  the  pipette  is  left,  and  is  not 
blown  out. 

131.  Burettes. — An  ordinary  burette  has  been  already 
described  in  Art.  118;  however,  a  few  practical  hints  may 
be  useful  to  the  student. 

If  the  figures  and  graduation  on  a  burette  are  not  easy  to 
read,  they  may  be  made  more  distinct  by  nibbing  them  over 
with  a  blue  pencil. 

The  surface  of  a  liquid  is  curved,  and  the  curve  is  called 
a  meniscus  ;  it  is  always  the  lowest  point  of  the  meniscus  that 
is  measured.  Note  the  following  points: 

The  eye  must  be  on  the  same  level  as  the  curve,  otherwise 
the  curve  is  seen  displaced  above  or  below  the  true  position, 
an  effect  known  as  parallax.  A  level  position  is  obtained  by 


104  THEORETICAL  CHEMISTRY.  §  5 

looking  straight  past  the  burette  at  some  point  known  to  be 
on  the  eye  level,  or  by  looking  past  the  curve  at  its  image  in 
a  looking  glass. 

The  extreme  edge,  or  lowest  position,  of  the  meniscus  is, 
however,  sometimes  hard  to  see,  but  it  can  easily  be  made 
clear.  If  the  background  is  bright,  the  meniscus  is  made 
dark  by  holding  a  dark  object  just  behind  and  below  the  sur- 
face. The  blackened  edge  of  a  card  slipped  over  the  burette 
tube  answers  admirably.  At  night,  however,  when  the  back- 
ground is  dark,  a  light  is  put  on  the  working  table  or  a  bench 
below,  to  illuminate  the  meniscus. 


A  SERIES 

OF 

QUESTIONS  AND  EXAMPLES 

RELATING  TO  THE  SUBJECTS 
TREATED  OF  IN  THIS  VOLUME. 


It  will  be  noticed  that  the  various  Question  Papers  that 
follow  have  been  given  the  same  section  numbers  as  the 
Instruction  Papers  to  which  they  refer.  No  attempt  should 
be  made  to  answer  any  of  the  questions  or  to  solve  any  of 
the  examples  until  the  Instruction  Paper,  having  the  same 
section  number  as  the  Question  Paper  in  which  the  questions 
or  examples  occur,  has  been  carefully  studied. 


ARITHMETIC. 

(ARTS.  1-181.     SEC.  1.) 


(1 )  What  is  arithmetic  ? 

(2)  What  is  a  number  ? 

(3)  What  is  the  difference  between  a  concrete  number 
and  an  abstract  number  ? 

(4)  Define  notation  and  numeration. 

(5)  Write  each  of  the  following  numbers  in  words : 

(a)  980;  (b)  605;  (c)  28,284;  (d)  9,006,042;  (e)  850,- 
317,002;  (/)  700,004. 

(6)  Represent  in  figures  the  following  expressions : 

(a)  Seven  thousand  six  hundred,  (b)  Eighty-one  thousand 
four  hundred  two.  (e)  Five  million  four  thousand  seven. 
(d)  One  hundred  eight  million  ten  thousand  one.  (e)  Eight- 
een million  six.  (/)  Thirty  thousand  ten. 

(7)  What   is   the   sum   of   3,290+504  +  865,403  +  2,074 
+  81  +  7?  Ans.   871,359. 

(8)  709  +  8,304,725  +  391  +  100,302  +  300  +  909  =  what? 

Ans.  8,407,336. 

(9)  Find  the  difference  between  the  following: 
(a)  50,962  and  8,3S8;  (b)  10,001  and  15,339. 

((a)     47,624. 
(  (b)     5,338. 


2  ARITHMETIC.  §  1 

(10)  (a)  70,968-32,975  =  ?     (b)  100,000-98,735  =  ? 

j  (*)     37,993. 
;'  \  (b)     1,265. 

(11)  The  greater  of  two  numbers  is  1,004  and  their  differ- 
ence is  49;  what  is  their  sum  ?  Ans.   1,959. 

(12)  From  5,962  +  8,471  +  9,023  take  3,874  +  2,039. 

Ans.   17,543. 

(13)  A  man  willed  $125,000  to  his  wife  and  two  children ; 
to  his  son  he  gave  $44,675,  to  his  daughter  $26,380,  and  to 
his  wife  the  remainder.     What  was  his  wife's  share  ? 

Ans.   $53,945. 

(14)  Find  the  products  of  the  following : 

(a)  526,387X7;  (b)  700,298x17;  (c)  217x103x67. 

(  (a)     3,684,709. 
Ans.  4  (b)     11,905,066. 
(  (c)     1,497,517. 

(15)  If  your  watch  ticks  once  every  second,  how  many 
times  will  it  tick  in  one  week  ?  Ans.  604, 800  times. 

(16)  If  a  monthly  publication  contains  24  pages  in  each 
issue,  how  many  pages  will  there  be  in  8  yearly  volumes  ? 

Ans.  2,304. 

(17)  An  engine  and  boiler  in  a  manufactory  are  worth 
$3,246.     The  building  is  worth  three  times  as  much,  plus 
$1,200,  and  the  tools  are  worth  twice  as  much  as  the  build- 
ing, plus  $1,875.     (a)  What  is  the  value  of  the  building  and 
tools  ?     (b)  What  is  the  value  of  the  whole  plant  ? 

Ans  I  (*>     $34'689' 
"  \  (b)     $37,935. 

(18)  Solve  the  following  by  cancelation: 
72X48X28X5        ?  80x60x50x16x14 

W    96X15X7X6    :  W       70X50X24X20 

A  J   W       8' 

Ans'  (  (b)     32. 


§  1  ARITHMETIC.  3 

(19)  If  a  mechanic  earns  $1,500  a  year  for  his  labor,  and 
his  expenses  are  $908  per  year,  in  what  time  can  he  save 
enough  to  buy  28  acres  of  land  at  $133  an  acre  ? 

Ans.  7  yr. 

(20)  A  freight  train  ran  365  miles  in  one  week,  and  3 
times  as  far,  lacking  246  miles,  the  next  week  ;  how  far  did 
it  run  the  second  week  ?  Ans.  849  mi. 

(21)  If  the  driving  wheel  of  a  locomotive  is  16  feet  in 
circumference,  how  many  revolutions  will  it  make  in  going 
from  Philadelphia  to  Pittsburg,  the  distance  between  which 
is  354  miles,  there  being  5,280  feet  in  one  mile  ? 

Ans.  116,820  rev. 

(22)  What  is  the  quotient  of: 

(*)  589,  824  -T-  576?     (b)  369,730,620-^-43,911?     (c)  2,527,- 
525-4-505?     (d)  4,  961,  794,  302  -^  1,234?         (  1,024. 

8,420. 


4,020,903. 

(23)  A  man  paid  $444  for  a  horse,  wagon,  and  harness. 
If  the  horse  cost  $264  and  the  wagon  $153,  how  much  did 
the  harness  cost  ?  Ans.  $27. 

(24)  What  is  the  product  of: 

(a)  1,024X576?     (b)  5,005x505?     (c)  43,911x8,420? 

(  (a)    589,824. 
AnsJ  (£)     2,527,525. 
(  (c)     369,730,620. 

(25)  If  a  man  receives  30  cents  an  hour  for  his  wages, 
how  much  will  he  earn  in  a  year,  working  10  hours  a  day 
and  averaging  25  days  per  month  ?  Ans.  $900. 

(26)  What  is  a  fraction  ? 

(27)  What  are  the  terms  of  a  fraction  ? 

(28)  What  does  the  denominator  show  ? 

(29)  What  does  the  numerator  show  ? 

(30)  How  do  you  find  the  value  of  a  fraction  ? 


4  ARITHMETIC.  §  1 

(31)  Is  -^-  a  proper  or  an  improper  fraction,  and  why  ? 

(32)  Write  three  mixed  numbers. 

(33)  Reduce  the  following-  fractions  to  their  lowest  terms  : 
t>  A*  A,  H-  Ans.  $,  i,  i,  |. 

(34)  Reduce  6  to  an  improper  fraction  whose  denomina- 
tor is  4.  Ans.  -2^. 

(35)  Reduce  7J,  13T\,  and  lOf  to  improper  fractions. 

Ans.  *£, 


(36)  What  is  the  value  of  each  of  the  following:  -1/,  -L7-, 
>  ¥>  H  ?  .  Ans. 

(37)  Solve  the  following: 

(a)    35- 
I5|~4f. 

Ans.       (c) 


(38)  i  +  |  +  f  =  ?  Ans. 

(39)  i  +  f+A  =  ?  Ans.  i 

(40)  424-31f  +  9^  =  ?  Ans. 


(41)  An  iron  plate  is  divided  into  four  sections;    the  first 
contains  29f  square  inches;  the  second,  50f  square  inches; 
the   third,  41   square  inches;    and  the  fourth,  69^-   square 
inches.     How  many  square  inches  are  in  the  plate  ? 

Ans.  190T9¥  sq.  in. 

(42)  Find  the  value  of  each  of  the  following  : 

16  4  +  3 


.       . 

Ans-     w  *. 

16  8  *W     A- 

(43)     The  numerator  of  a  fraction  is  28,  and  the  value  of 
the  fraction  J  ;  what  is  the  denominator  ?  Ans.  32, 


§  1  ARITHMETIC.  5 

(44)  What  is  the  difference  between  (a)  J-  and  T^  ?  (b)  13 
and  7^  ?    (c)  312^  and  229¥\  ? 

((«)    TV 

Ans.  -{  (*)     5TV 

((c)     83H- 

(45)  If  a  man  travels  85T\  miles  in  one  day,  78^  miles  in 
another  day,  and  125|~J  miles  in  another  day,  how  far  did  he 
travel  in  the  three  days  ? 

Ans.  289ffJ-  mi. 

(46)  From  573f  tons  take  216f  tons.  Ans.  357^  T. 

(47)  At  f  of  a  dollar  a  yard,  what  will  be  the  cost  of  9£ 
yards  of  cloth  ?  Ans.  3^-f  dollars. 

(48)  Multiply  f  of  f  of  ^  of  |f  of  11  by  £  of  £  of  45. 

Ans. 


(49)  How  many  times  are  f  contained  in  f  of  16  ? 

Ans.  18  times. 

(50)  Bought  211^  pounds  of  old  lead  for  1£  cents  per 
pound.     Sold  a  part  of  it  for  2^  cents  per  pound,  receiving 
for  it  the  same  amount  as  I  paid  for  the  whole.     How  many 
pounds  did  I  have  left  ? 

Ans.  52      Ib. 


(51)  Write   out   in   words  the   following  numbers:  .08, 
.131,  .0001,  .000027,  .0108,  and  93.0101. 

(52)  How  do  you  place  decimals  for  addition  and  sub- 
traction ? 

(53)  Give  a  rule  for  multiplication  of  decimals. 

(54)  Give  a  rule  for  division  of  decimals. 

(55)  State  the  difference  between  a  fraction  and  a  decimal. 

(56)  State  how  to  reduce  a  fraction  to  a  decimal. 

1-32 


ARITHMETIC.  §  1 

(57)     Reduce  the  following  fractions  to  equivalent  deci- 


mals: |,  J,  A,  T™  and 


Ans. 


.5. 

.875. 

.15625. 


.65. 
[  .125. 
(58)     Solve  the  following: 

32.5 +.29 +1.5  ,     1.283X8+1* 


Ans.  - 


589  +  27X163-8.  40.6 +  7.1  X  (3.029 -1.874) 

25  +  39  6.27  +  8.53-8.01 

i     2.5029. 
6.3418. 
1,491.875. 
)    8.1139. 

(59)  How  many  inches  in  .875  of  a  foot  ?        Ans.  10^-  in. 

(60)  What  decimal  part  of  a  foot  is  T3^  of  an  inch  ? 

Ans.  .015625. 

(61)  A  cubic  inch  of  water  weighs  .03617  of  a  pound. 
What  is  the  weight  of  a  body  of  water  whose  volume  is 
1,500  cubic  inches  ?  Ans.   54.255  Ib. 

(62)  If  by  selling  a  carload  of  coal  for  $82. 50,  at  a  profit 
of  $1.65  per  ton,  I  make  enough  to  pay  for  72.6  feet  of 
fencing  at  $.  50  a  foot,  how  many  tons  of  coal  were  in  the 
car?  Ans.   22  T. 

(63)  Divide  17,892  by  231,  and  carry  the  result  to  four 
decimal  places.  Ans.   77.4545+. 

(64)  What  is  the  value  of  the  following  expression  car- 
ried to  three  decimal  places : 

74. 26  X  24  X  3. 1416  X  19  X  19  X  350 


33,000X12X4 


=  ?    Ans.  446. 619- . 


(65)  Express:  (a)  .7928  in  64ths;  (b)  .1416  in  32ds; 
(c)  .47915  in  16ths.  f  (a)  fi- 

Ans.     (b)     A. 

IW    A- 


§1 


ARITHMETIC. 


(66)     Work  out  the  following  examples: 
(a)    709. 63 -.8514;     (b)    81.963-1.7;     (c) 

1-.001;     (e)     872.1 -(.8721 +  .008);      (/) 

-(6.704-2.38). 


Ans.  - 


18-. 18;    (d) 
(5. 028 +  .0073) 

(a)  708.7786. 

(b)  80.263. 
17.82. 
.999. 
871.2199. 
.7113. 


w 

(d) 

w 

I  (/) 


(67)     Work  out  the  following: 

(a)  £-.807;  (b)  .875 -f;  (<:)  (A +  -435)  -  (AV--07); 
(^/)  What  is  the  difference  between  the  sum  of  33  millionths 
and  17  thousandths,  and  the  sum  of  53  hundredths  and  274 


thousandths  ? 


Ans. 


(a)  .068. 

(b)  .5. 

(c)  .45125. 

(d)  .786967. 


(68)     What  is  the  sum  of  .125,   .7,   .089,  .4005,  .9,   and 
.000027?  Ans.   2.214527. 


(69)  927.416  +  8.274  +  372.6  +  62.07938  =  ? 

Ans.  1,370.36938. 

(70)  Add  17  thousandths,  2  tenths,  and  47  millionths. 

Ans.   .217047. 

(71)  Find  the  products  of  the  following  expressions  : 

(a)  .013  X.  107;  (b)  203x2.03  X.  203;  (c)  2.7x31.85x  (3.16 
-.316);  (d)  (107.8  +  6.541-  31.96)  X  1.742. 

(a) 


Ans. 


(b) 
(c) 
(d) 


.001391. 
83.65427. 
244.56978. 
143.507702. 


(72)     Solve  the  following 
(a)  (A  -•  13)  X  1625+1;  ( 
+  .013-  2.17)  X13J-7A- 


(«X.21)  -(.02X  A);  (*)  « 

f  («)     -384375. 
Ans.  J  (b)     .1209375. 
I  (c)     6.4896875. 


ARITHMETIC.  §  1 

(73)  Solve  the  following : 

(a)  .875-4-i;  (*)  |^.5;  M^^Jr 

T^  — .1^5  r  (#)     1.75, 

Ans.  J  (^)     1.75. 
1(0     -5. 

(74)  Find  the  value  of  the  following"  expression : 

1.25X20X3 
87  + (11X8)' 
459  +  32  Ans.  210f 

(75)  From  1  plus  .001  take  .01  plus  .000001. 

Ans.  .990999. 


ARITHMETIC. 

(ARTS.  1-168.     SEC.  2.) 

(1)  What  is  25  per  cent,  of  8,428  Ib.  ?  Ans.  2,107  Ib. 

(2)  What  is  1  per  cent,  of  $100  ?  Ans.   $1. 

(3)  What  is  1  per  cent,  of  $35,000  ?  Ans.  $175. 

(4)  What  per  cent,  of  50  is  2  ?  Ans.  4$. 

(5)  What  per  cent,  of  10  is  10  ?  Ans.   100$. 

(6)  Solve  the  following: 

(a)  Base  =  $2,522  and  percentage  =  $176.54.  What  is 
the  rate  ?  (b)  Percentage  =  16.96  and  rate  =  8  per  cent. 
What  is  the  base?  (c)  Amount  =  2 16. 7025  and  base  =  213.5. 
What  is  the  rate?  (d)  Difference  =  201.825  and  base 
=  207.  What  is  the  rate  ? 


(b)      212. 

(c) 
(d) 


Ans. 


(7)  A  farmer  gained  15$  on  his  farm  by  selling  it  for 
$5,500.     What  did  it  cost  him  ?  Ans.  $4,782.61. 

(8)  A  man  receives  a  salary  of  $950.     He  pays  24$  of  it 
for  board,  12^-$  of  it  for  clothing;  and  17$  of  it  for  other 
expenses.     How  much  does  he  save  in  a  year  ?    Ans.  $441. 75. 

(9)  If  37^  per  cent,  of  a  number  is  961.38,  what  is  the 
number?  Ans.   2,563.68. 


2  ARITHMETIC.  §  2 

(10)  A  man  owns  £  of  a  property.     30$  of  his  share  is 
worth  $1,125.     What  is  the  whole  property  worth  ? 

Ans.  $5,000. 

(11)  What  sum  diminished  by  35$  of  itself  equals  $4,810  ? 

Ans.  $7,400. 

(12)  A  merchant's  sales  amounted  to  $197.55  on  Monday, 
and  this  sum  was  12|$  of  his  sales  for  the  week.    How  much 
were  his  sales  for  the  week  ?  Ans.  $1,580.40. 

(13)  The  distance  between  two  stations  on  a  certain  rail- 
road is  16.5  miles,  which  is  12-j-$  of  the  entire  length  of  the 
road.    What  is  the  length  of  the  road  ?  Ans.   132  mi. 

(14)  After  paying  60$  of  my  debts  I  find  that  I  still  owe 
$35.     What  was  my  whole  indebtedness  ?  Ans.  $87.50. 

(15)  Reduce  28  rd.  4  yd.  2  ft.  10  in.  to  inches. 

Ans.   5, 722  in. 

(16)  Reduce  5,722  in.  to  higher  denominations. 

Ans.   28  rd.  4  yd.  2  ft.  10  in. 

(17)  How  many  seconds  in  5  weeks  and  3.5  days  ? 

Ans.  3,326,400  sec. 

(18)  How   many    pounds,    ounces,     pennyweights,    and 
grains  are  contained  in  13,750  gr.  ? 

Ans.   2  Ib.  4  oz.  12  pwt.  22  gr. 

(19)  Reduce  4,763,254  links  to  miles. 

Ans.   595  mi.  32  ch.  54  li. 

(20)  Reduce  764,325  cu.in.  to  cu.yd. 

Ans.   16  cu.yd.  10  cu.ft.  549  cu.in. 

(21)  What  is  the  sum  of  2  rd.  2  yd.  2  ft.  3  in. ;  4  yd.  1  ft. 
9  in. ;  2  ft.  7  in.?  Ans.   3  rd.  2  yd.  2  ft.  1  in. 

(22)  What   is  the  sum  of  3  gal.  3  qt.  1  pt.  3  gi. ;    6  gal. 
1  pt  2  gi. ;  4  gal.  1  gi. ;  8  qt.  5  pt.  ?     Ans.   16  gal.  3  qt.  2  gi. 

(23)  What  is  the  sum  of  240  gr.  125  pwt.  50  oz.  and  3  Ib.  ? 

Ans.  7  Ib.  8  oz.  15  pwt. 


§  2  ARITHMETIC.  3 

(24)  What  is  the  sum  of  11°  16'  12";    13°  19'  30";  20° 
25";  26'  29";  10°  17'  11"  ?  Ans.  55°  19'  47". 

(25)  What  is  the  sum  of  130  rd.  5  yd.  1  ft.  6  in. ;  215  rd. 

2  ft.  8  in. ;  304  rd.  4  yd.  11  in.  ?    Ans.  2  mi.  10  rd.  5  yd.  7  in. 

(26)  What  is  the  sum  of  21  A.  67  sq.ch.  3  sq.rd.  21  sq.li. ; 
28  A.  78  sq.ch.  2  sq.rd.  23  sq.li.;  47  A.  6  sq.ch.  2  sq.rd. 
18  sq.li.  ;  56  A.  59  sq.ch.  2  sq.rd.  16  sq.li.  ;  25  A.  38  sq.ch. 

3  sq.rd.  23  sq.li. ;  46  A.  75  sq.ch.  2  sq.rd.  21  sq.li.? 

Ans.  255  A.  3  sq.ch.  14  sq.rd.  122  sq.li. 

(27)  From  20  rd.  2  yd.  2  ft.  9  in.  take  300  ft. 

Ans.  2  rd.  1  yd.  2  ft.  9  in. 

(28)  From  a  farm  containing  114  A.  80  sq.rd.  25  sq.yd., 
75  A.  70  sq.rd.  30  sq.yd.  are  sold.     How  much  remains  ? 

Ans.   39  A.  9  sq.rd.  25£  sq.yd. 

(29)  From  a  hogshead  of  molasses,  10  gal.  2  qt.  1  pt.  are 
sold  at  one  time,  and  26  gal.  3  qt.  at  another  time.     How 
much  remains  ?  Ans.  25  gal.  2  qt.  1  pt. 

(30)  If  a  person  were  born  June  19,  1850,  how  old  would 
he  be  August  3,  1892  ?  Ans.  42  yr.  1  mo.  14  da. 

(31)  A  note  was  given  August  5,   1890,   and   was  paid 
June  3,  1892.     What  length  of  time  did  it  run  ? 

Ans.   1  yr.  9  mo.  28  da. 

(32)  What  length  of    time    elapsed  from  16  min.   past 
10  o'clock  A.  M.,  July  4,  1883,  to  22  min.  before  8  o'clock  p.  M., 
Dec.  12,  1888  ?  Ans.  5  yr.  5  mo.  8  da.  9  hr..22  min. 

(33)  If  1  iron  rail  is  17  ft.  3  in.  long,  how  long  would 
51  rails  be,  if  placed  end  to  end  ?         Ans.   53  rd.  1|  yd.  9  in. 

(34)  Multiply  3  qt.  1  pt.  3  gi.  by  4.7. 

Ans.  4  gal.  2  qt.  1.7  gi. 

(35)  Multiply  3  Ib.  10  oz.  13  pwt.  12  gr.  by  1.5. 

Ans.  5  Ib.  10  oz.  6  gr. 


4  ARITHMETIC.  §  2 

(36)  How  many  bushels  of  apples  are  contained  in  9  bbl., 
if  each  barrel  contains  2  bu.  3  pk.  6  qt.  ? 

Ans.  26  bu.  1  pk.  6  qt. 

(37)  Multiply  7  T.  15  cwt.  10.5  Ib.  by  1.7. 

Ans.   13  T.  3  cwt.  67.85  Ib. 

(38)  Divide  358  A.  57  sq.rd.  6  sq.yd.  2  sq.ft.  by  7. 

Ans.   51  A.  31  sq.rd.  8  sq.ft. 

(39)  Divide  282  bu.  3  pk.  1  qt.  1  pt  by  12. 

Ans.   23  bu.  2  pk.  2  qt.  i  pt. 

(40)  How  many  iron  rails,  each  30  ft.  long,  are  required 
to  lay  a  railroad  track  23  mi.  long  ?  Ans.   8,096  rails. 

(41)  How  many  boxes,   each  holding  1  bu.  1  pk.  7  qt., 
can  be  filled  from  356  bu.  3  pk.  5  qt.  of  cranberries  ? 

Ans.  243  boxes. 

(42)  If  16  square  miles  are  equally  divided  into  62  farms, 
how  much  land  will  each  contain  ? 

Ans.   165  A.  25  sq.rd.  24  sq.yd.  3  sq.ft.  80+  sq.in. 

(43)  What  is  the  square  of  108  ?  Ans.   11,664. 

(44)  What  is  the  cube  of  181.25  ?  Ans.   5,954,345.703125. 

(45)  What  is  the  fourth  power  of  27.61  ? 

Ans.  581,119.73780641. 

(46)  Solve  the  following:  (#)1062;  (b)  (182i)2;  (c)  .0052; 


(d)  .00632;  (e)  10. 062. 


Ans.  -i 


(a)  11,236. 

(b)  33,169.515625. 

(c)  .000025. 

(d)  .00003969. 

(e)  101.2036. 


(47)     Solve  the  following:  (a)  7533;  (b)  987. 43;  (c)  .0053; 


(d)  .40443. 


Ans.  - 


(a)  426,957,777. 

(&)  962,674,279.624. 

(c)  .000000125. 

(d)  .066135317184. 


§  2                                  ARITHMETIC.  5 

(48)  What  is  the  fifth  power  of  2  ?  Ans.   32. 

(49)  What  is  the  fourth  power  of  3  ?  Ans.  81. 

(50)  What  are  the  values  of:  (a)  67.852?  (b)  967,8452? 

(a)     4,603.6225. 


Ans.  - 


(b)      936,723,944,025. 
W      A- 


(51)  What  is  (a)  the  tenth  power  of  5  ?   (b)  the  fifth  power 
of  9?  \(a)     9,765,625. 

'   (  (b)     59,049. 

(52)  Solve  the  following:  (a)  1.24;  (b)  II6;  (c)  I7;  (d}  .Ol4; 
W  -1&-  [  (a)     2.0736. 

(b)  1,771,561. 

Ans.  -    (V)  1. 

(d)  .00000001. 

.  (e)  .00001. 

(53)  Find  the  values   of   the  following:    (a)  .01333;    (b) 


301.0113;  (c)  (i)3;  (d)  (3f)3. 

Ans. 


(a)  .000002352637. 

(b)  27,  273,  890.  942264331. 


(d)     52££,  or  52.734375. 


(54)  In  what  respect  does  evolution  differ  from  involution  ? 

NOTE. — In  the  answers  to  the  following  examples,  a  minus  sign  after 
a  number  indicates  that  the  last  digit  is  not  quite  as  large  as  the  num- 
ber printed.  Thus,  12.497—  indicates  that  the  number  really  is  12.496+ , 
and  that  the  6  has  been  made  a  7  because  the  next  succeeding  figure 
was  5  or  greater.  For  example,  had  it  been  desired  to  use  but  three 
decimal  places  in  example  46  (<£),  the  answer  would  have  been  written 
33,169.516-. 

(55)  Find  the  square  root  of  the  following:  (a)  3,486,- 
784.401;  (b)  9,000,099.4009;  (c)  .001225. 

[(a)     1,867.29+. 
Ans.  I  (b)     3,000.017-. 
I  (c)     .035. 


G 


ARITHMETIC. 


(56)     Extract  the  square  root  of  (a)  10,  795.  21;  (b)  73,008.04; 
(c)  90;  (d)  .09.  (  (a)      103.9. 

b       270.2. 


(d)     .3. 

(57)     Extract  the  cube  root  of  (d)  .32768;  (£)  74,088;  (c) 
92,416;  (rf)  .373248.  (  (a)      .6894+. 


(,)      45.212-. 
(</)     .72. 

(58)  Extract  the  cube  root  of  2  to  six  decimal  places. 

Ans.   1.259921+  . 

(59)  Extract    the    cube    root    of    (a)    1,758.416743;    (b) 
1,191,016;  (c)  A;  (d)  AV  (a)      12.07. 

b       106. 


(60)     Extract  the  cube  root  of  3  to  six  decimal  places. 

Ans.   1.  442250-  . 


'502,681;  (d)  ^.00041209. 


(61)     Solve  the  following:  (a)  V123.21;  (b)  V114.921;  (c) 

(a)  11.1. 

(b)  10.72+. 
y      709. 
(d\     .0203. 


(62)     Solve    the    following:    (a)  ^.0065;    (b)  f.021;    (c) 
1^8,036,054,027;  (d)  ^.000004096;  (e)  $17. 

(a)  .18663-. 

(b)  .2759-. 
Ans.  4  (c)  2,003. 

(d)  .016. 

(e)  2.5713-. 


ARITHMETIC. 


(63)     Solve  the  following:  (a)   i/  0,501;  (l>)   fl!7,649;  (c) 
.-000064;  (d)  ff.  f   a}     9. 


(</)     .72112+. 


(64)     Extract   the   square  root  of:    (a)  Hrf J    (&)   -3364; 
(c)  .1;  (d)  25.0|;  (e)  .OOOf. 

.58. 


Ans.  •< 


(<:)      .31623-. 

(d)  5. 00749+ . 

(e)  .02108+. 


(65)  (a)  Extract  the  fourth  root  of  2  to  four   decimal 
places,      (b)  Extract  the  sixth  root  of  6. 

((a)     1.1892+. 
''\(b)     1.  34801-  . 

(66)  Extract  the  square  root  of  (a)  3.1416  and  (b)  .7854 
to  four  decimal  places.  j  (a)     1.7725—. 

M(J)     .8862+. 

(67)  Extract  the  cube  root  of  (a)  3.1416  and  (b)  .5236  to 
four  decimal  places.  j  (#)     1.4646—. 

''{(b)     .8060-. 

Find  the  value  of  x  in  the  following  : 

(68)  11.  7:  13::  20:*.  Ans.  22.22+. 

(69)  (0)20  +  7:10  +  8::3:;r;  (b)  122  :  1002  ::  4  \x. 


'  x 
W_       jx^_ 
150  ~  600* 


16 


Ans 
L 


10  =  I00 


.277.7+. 
16        60 


Ans.  -< 


(*)  *  =  12. 

(^)  x  =  12. 

(V)  *  =  20. 

(d)  x  =  180. 

(e)  x  =  40. 


8                                      ARITHMETIC.  §  2 

(71)  x\ 5::  27: 12.5.  Ans.  lOf 

(72)  45:  60::*:  24.  Ans.  18. 

(73)  *:35::4:7.  Ans.  20. 

(74)  9:*::  6:  24.  Ans.  36. 

(75)  ^1,000:^331  =  27:*.  Ans.  29.7. 

(76)  64:81  =  212:*2.  Ans.  23.625. 

(77)  7  +  8:7  =  30:*.  Ans.   14. 

(78)  A  man  whose  steps  measure  2  ft.  5  in.  takes  2,480 
steps  in  walking  a  certain  distance.     How  many  steps  of  2  ft. 
7  in.  will  be  required  for  the  same  distance  ? 

Ans.   2,320  steps. 

(79)  If  a  horse  travels  12  mi.  in  1  hr.  36  min.,  how  far 
will  he  travel  at  the  same  rate  in  15  hr.  ?          Ans.   112.5  mi. 

(80)  If  a  column  of  mercury  27.63  in.  high  weighs  .76  of 
a  pound,  what  will  be  the  weight  of  a  column  of  mercury 
having  the  same  diameter,  29.4  in.  high  ?         Ans.   .808+  Ib. 

(81)  If  2  gal.  3  qt.  1  pt.  of  water  will  last  a  man  5  da.,  how 
long  will  5  gal.  3  qt.  last  him,  if  he  drinks  at  the  same  rate  ? 

Ans.   10  da. 

(82)  Heat  from  a  burning  body  varies  inversely  as  the 
square  of  the  distance  from  it.     If  a  thermometer  held  6  ft. 
from  a  stove  shows  a  rise  in  temperature  of  24°,  how  many 
degrees  rise  in  temperature  would  it  indicate  if  held  12  ft. 
from  the  stove  ?  Ans.   6°. 

(83)  If  a  pile  of  wood  12  ft.  long,  4  ft.  wide,  and  3  ft.  high 
is  worth  $12,  what  is  the  value  of  a  pile  of  wood  15  ft.  long, 
5  ft.  wide,  and  6  ft.  high?  Ans.  $37.50. 

(84)  If  100  gal.  of  water  run  over  a  dam  in  2  hr.,  how 
many  gallons  will  run  over  the  dam  in  14  hr.  28  min.  ? 

Ans.  723J  gal. 


§  2  ARITHMETIC.  9 

(85)  If  a  cistern  28  ft.  long,  12  ft.  wide,  10  ft.  deep  holds 
798  bbl.  of  water,  how  many  barrels  of  water  will  a  cistern 
hold  that  is  20  ft.  long,  17  ft.  wide,  and  G  ft.  deep  ? 

Ans.  484|  bbl. 

(86)  If  a  railway  train  runs  444  mi.  in  8  hr.  40  min.,  in 
what  time  can  it  run  1,060  mi.  at  the  same  rate  of  speed  ? 

Ans.  20  hr.  41.44  min. 

(87)  If  sound  travels  at  the  rate  of  6,160  ft.  in  5£  sec., 
how  far  does  it  travel  in  1  min,?  Ans.   67,200  ft. 

(88)  If  5  men  by  working  8  hours  a  day  can  do  a  certain 
amount  of  work,  how  many  men  by  working  10  hours  a 
day  can  do  the  same  work  ?  Ans.   4  men. 

(89)  If  a  man  travels  540  miles  in  20  days  of  10  hours 
each,  how  many  hours  a  day  must  he  travel  to  cover  630 
miles  in  25  days  ?  Ans.   9^  hr. 

(90)  Referring  to  example  4,  Art.  168,  Arithmetic,  §  2, 
what  is   the    horsepower  of   an  engine  whose  cylinder  is 
30   inches  in  diameter,  piston  speed,  660  feet  per  minute, 
and  mean  effective  pressure,  42  pounds  per  square  inch  ? 

Ans.   594  horsepower. 

(91)  The   weight  of   a  cubic   inch   of  cast   iron   is  .261 
pound.     Referring   to  Art.  164,  Arithmetic,   §  2,  what  is 
the  weight  of  a  solid  cast-iron  cylinder  whose  diameter  is 
12  inches  and  length  is  60  inches  ?  Ans.   1,771.11  Ib. 

(92)  Referring   to  Art.    167,  Arithmetic,  §  2,   what   is 
the   centrifugal  force   of  a  40-pound  body  revolving  in  a 
circle  having  a  radius  of  10  inches,  at  a  speed  of  18  feet  per 
second  ?  Ans.   484. 7  Ib. 


ELEMENTARY  ALGEBRA 


AND 


TRIGONOMETRIC  FUNCTIONS. 


(1)  How  may  the  signs  of  all  the  terms  of  the  denomi- 
nator of  a  fraction  be  changed  from  -f-  to  —  or  from  —  to  -f- 
without  altering  the  value  of  the  fraction  ? 

(2)  Solve  the  equations: 

(a)  3^  +  6-2^  =  7*;   (b)  5* -(3* -7)  =  ±x  -  (6*  -  35) ; 
(c)   (-r  +  5)2-(4-;t-)2  =  21*.  f  (a)     x  =  1. 

Ans.    i  (b)     x  =  7. 
(  (c)     x  =  3. 

(3)  Divide: 

30s  +  2  -  4rt5  +  7«  +  Za°  —  5a*  + 10^2  by  a3  —  1  -  a2  —  2a. 

Ans.   2<73  —  2«2  —  3^  —  2. 

(4)  Multiply : 

(0)     2  +  40  -  5rt2  -  Gtf 3  by  7rt3 ;    (/?)  4^2  -  4/  +  Gs2  by  3^r2j/ ; 
(c)   3£  +  5<:-2</by  60. 

r  (0)     14^3  +  28rt4  -  350*  -  42^fi. 
Ans.   J  (£)     12^4j/-12^y 
I  (^-)     18^^  +  30^  - 

(5)  A  man  performed  a  journey  of  48  miles  in  a  certain 
number  of  hours ;  but  if  he  had  traveled  4  miles  more  each 
hour,  he   would  have  performed  the  journey  in   6   hours. 
How  many  miles  did  he  travel  per  hour  ?  Ans.  4  miles. 

§  3 


2  ELEMENTARY  ALGEBRA  AND  §  3 

(6)     Translate    the  following-   algebraic   expressions  into 
ordinary  language : 


(7)     Find  the  products  of: 

y 

; 

Ans. 


+  16.  « 

(8)  Find  the  sine,  cosine,  and  tangent  of  17°  27'  37". 

(9)  A  vessel  containing  some  water  was  filled  by  pouring 
in  42  more  gallons;   there  was  then   seven  times  as  much 
water  in  the  vessel  as  at  first.     How  much  did  the  vessel 
hold  ?  Ans.   49  gallons. 


(10)  Reduce  to  its  simpiest  form. 

(a  +  b}c 

(11)  Find  the  sine,  cosine,  and  tangent  of  63°  4'  51.8". 

(12)  Find  the  values  of  the  following: 


v,  *      2-3*  .   \§x-x* 

(13)     Simplify-  __  +  _^  Ans. 


cf  +  f  —  b1-  2ac 
(14)     Multiply    ,  by  . 


SUGGESTION. — Factor  the  numerators  and  denominators  before  mul- 
tiplying. 

.  -  a  +  b  —  c 

Ans.  — — -— j— — r-. 
ac(a  —  b-\-c] 

(15)     From  a*  —  b*  take  §a*b  —  la* ft*  +  5ab3,  and  from  the 
result  take  3# 4  —  ±a*b  +  6^2^2  +  bab*  —  3^4. 
Ans.   —  2tf4  — 


(16)     Sine  =  .27038,  cosine  =  .27038,  and  tangent 
=  2.27038;  find  the  corresponding  angles. 


§  3  TRIGONOMETRIC  FUNCTIONS.  3 

(17)  (a)  Give  an  illustration,  not  contained  in  the  section 
on  Algebra,  that  will  explain  the  difference  between  positive 
and  negative  quantities,      (b)  In  what  respects  are  addition 
and  subtraction  different  in  algebra  from  addition  and  sub- 
traction in  arithmetic  ? 

(18)  (a)    What    is    the    value    of    a°  ?      (b)    What    does 
a°  -+-  a~l  equal  ? 

(19)  Change  the  fraction  —C~  /^  r-r!    so    that    the    ^g11 
before  the  dividing  line  will  be  -(-. 

(20)  (a)    What  is   the   reciprocal  of  f  f  ?      (b)    Of   what 
number  is  700  the  reciprocal  ? 

(21)  (a)  From  3a  —  2b  +  3c  take  %a  —  lb  —  c  —  b.    (b)  Sub- 
tract     x*+y*  —  xy*    from    2*3  —  3*>  +  &try.       (<:)    From 
14*  +  4£  -  Qc  -  3d  take  110  -  2£  +  4^  -  4*/. 


Ans.  i   (b)     x3  —  %x*y  -f-  xy 
I  (c)     3 


(22)     Find  the  numerical  values  of  the  following  expres- 
sions when  a  =  16,  b  =  10,  and  x  =  5  :    (#)i  (ab^x  -f- 


614,400. 
Ans.      (*)     i     ' 

(<:)     23,400. 

(23)  Find  the  sum  of  the  following  :  (a)  ^xyz  —  Sxyz  —  5xyz, 
Qxyz  —  §xyz  +  Zxyz-,    (b)    3«2  +  lab  +  4 

-a*  +  5ab-b\    18«2  -  20^  -  19£2,    and 

(<;)  4ww  +  3^  —  4<r,  3^  —  4^  +  2;^,  and3;^2  —  4/. 

f  (0)      -±xyz. 
Ans.  j  (^) 

I  \c] 

(24)  (a)  Explain    in    your    own    words    the   difference 
between  a  coefficient  and  an  exponent,     (b}  How  are  coeffi- 
cients and  exponents  treated  in  multiplication,  and  how  in 
division  ?     (c)  What  is  the  law  of  signs  in  multiplication  ? 


4  ELEMENTARY  ALGEBRA  AND  §  3 

(25)  In  a   right    triangle   ABC,    the    hypotenuse   A  B 
=  17.69  feet,  and  the  side  A  C  =  9  f  t.  9  in.  ;  find  the  other 
three  parts.  f  56°  33'  15". 

Ans.  I  33°  26'  45". 
I  14  ft.  9  in. 

(26)  Solve  the  equations  : 


ax      x  (  (a)     x  —  8. 

3 


bm. 


(27)     State  how  you  would    read    the  following  expres- 
sions : 


(a)  a'**  +  2rt'£B  -  (a  +  0)  ;  (b) 
(c)  (m  +  n)  (///_«)*    ;«_-    Y 


(28)  (a)  Write  a  monomial;  a  binomial;    a  polynomial. 
(#)  In  the  expression  a  -f-  20#  —  ^2,  why  cannot  the  indicated 
addition  and  subtraction  be   performed  ?     (c)  What  opera- 
tion is  indicated  between  the  quantities  in 

(29)  Resolve  into  their  factors  : 

(a)  45^7j/10-90^y  —  360^Y;   (b)  cftf 
(c)  (a+&y-(c-<t): 

Ans.   (c)  (a  +  b  +  c-d)  (a  +  b- 


(30)     Remove  the  symbols  of  aggregation  from  the  fol 
lowing  : 

(a)  Za  -  {  3/7  -f  [4<r  -  4^  -  (2a  +  2^)]  +  [3*  -  (^  +  0]  }  I 


f  («)     5a 

Ans.  ]  (t>)     5a. 

I    r       3« 


§  3  TRIGONOMETRIC  FUNCTIONS.  5 

(31)     In  a  right  triangle  A  B  C,  right-angled  at   C,  side 
A  C  =  17.5,  side  B  C  =  21.3;  find  the  other  three  parts. 

f  39°  24'  23". 
Ans.  I  50'  35'  37". 
I  27.  57,  nearly. 


(32)     In  the  formula,  t  =  I"  *,    it  is  required 

rr  ,£,  -f-  Vr^S^ 

to  transform  the  formula  so  that  /2  will  stand  alone  in  the 
first  member.     In  other  words,  solve  for  /2. 

Ans  ,  .. 


(33)  A  can  do  a  piece  of  work  in  5  days,  B  in  6  days, 
and  (C  in  7-J  days;  in  what  time  will  they  do  it,  working 
together  ?  Ans.  2  days. 


(34)     Simplify :     (a) 


;  + 


ab 


(') 


x+-  Ans.  < 


Z-x 


X. 

a 


b. 


b 
4 


(35)  («)  Arrange  a*t>*  +  2«^  +  3  -  7^2^3  +  6#4£4  according 
to  the  decreasing  powers  of  a  ;  (b)  according  to  the  increas- 
ing powers  of  b.  (c)  With  a?  -\-  1  +  2a3  -\-  ax  arranged  accord- 
ing to  the  increasing  powers  of  a,  should  the  1  be  placed  first 
or  last,  and  why  ? 


(36)     (a)  Express  with  radical  signs  :  .**;  3,r*j-i; 
(b)  Clear  a~lfa-\  --  -?-{-(m  —  n)~l          _3     of  negative  expo- 
nents.      (c)  Express  with  fractional  exponents: 


6  ELEMENTARY  ALGEBRA  AND  §  3 

(37)     Divide : 

(.       &CIX  — J—  X    .,              X  ,,.  \)TJl  ft    —  ofl  i  oil 

d]    — - — ! — —  by  ;  (b)        4  a — - — a  by       4  a  —  ; 

(«)     ^ 


;tr  - 


Ans.  -< 


(^)     2w'w  + 1. 


(38)     Multiply: 

\     /  I  *  ^  '      \     / 

and  (<:)  —  a3  -f-  3#2$  —  2^3  by  5^2 

f(«)      24r4- 
Ans.  j  (b)     2x*  — 

[(c)     -5tf5+6#4 


by 


(39)     Perform  the  indicated  additions: 

X         .    X  —  V  X*  X  X 


x-y     y-x 


+— nr-- 


Ans.  - 


(a) 
(*) 


84 


(40)     Factor       the      following:      (a)    9;r4  +  VZx^f  +  4j 


(41)     Divide  : 


9^3-l  by  3^-1; 


—  3. 


f  (a)     3x* 


Ans. 


(42)     Why  are  letters  used  in  algebra,  and  in  what  ways 
do  they  differ  from  figures  ? 


3  TRIGONOMETRIC  FUNCTIONS.  7 

(43)     (a)  Reduce  1  +  2,r ^—  to  a  fractional  form. 

3  f 2   I   2jtr  I    1 
(£)  Change  -  -  to  a  mixed  quantity. 


r       io.M^+4 

A  O^V 

Ans.  <! 

41 


(44)  In  a  right  triangle  ^  #  (7,  angle  y4  =  65°  13'  29", 
hypotenuse  A  B  =  5^-  yards ;    find    the    other  three  parts. 
Give  sides  in  feet  and  inches.  f  24°  46'  31/'. 

Ans.  xj  14  ft  1  If  in. 

I  6ft.  11  in.,  nearly. 

(45)  What    is    the    ratio    of    (a)   x*  —  1   to   x*  -f  1  ?     (£) 
^  _  2^f  +y*  to  4:  -y  ?  Ans.   (^)  (^2  - /)  ( 

(46)  Solve  the  following: 


"  ,r'-4' 
(a)     x  —  If 
Ans.  ^  (b)     x  =  2. 

(47)  Add  159°  27'  34.6",  25°  16'  8.7",  and  3°  48/  53". 

(48)  Combine  the  like  terms  of  the  expressions: 

f  («)    ST. 

Ans.  j  (^)     ^r  -  2^-. 
I  (c)      -  5a. 

(49)  Divide  :     (a)    3&m*y  +  28m*y*  —  l&tty3     by  —  litty  ; 
(<^)  4rt;4  —  3abb  —  a6b*  by  ^4 ;  (c)  4,r3  —  8^r5  +  12.r7  —  16j;9  by  4^2. 

r  (^^     _  5///2  _  &ny  -f-  2j/2. 
Ans.  -I  (b\     ^  —  'Sab—cfb*. 


8 


EL.  ALGEBRA  AND  TRIG.  FUNCTIONS.   §  3 


(50)  A  post  has  |  of  its  length  in   the  earth,  f  in  the 
water,  and  13  feet  in  the  air.     What  is  its  length  ? 

Ans.  35  feet. 

(51)  In  the  composition  of  a  quantity  of  gunpowder,  the 
niter  was  10  Ib.  more  than  J  of  the  whole,  the  sulphur  was 
4|-  Ib.  less  than  £  of  the  whole,  and  the  charcoal  was  2  Ib. 
less  than  |  of  the  niter.     What  was  the  amount  of  gun- 
powder and  of  each  of  the  ingredients  ? 

f  Gunpowder,  69  Ib. 

Ans.      Niter>  56  lb' 
Sulphur,  7  lb. 

Charcoal,  6  lb. 


(52)     Find  the  values  of  the  following: 

v;    (b)  m, 


(a)      -1 


^M;     W 

'  (a)      -5^/V. 
(b)      ±  lOaW 


Ans.  4 


(c)      3;«V. 


(53)  In  a  triangle  ABC,  side  A  B  =  70  feet,  side  B  C 
=  42  feet,  and  angle  A  =  36°  10'.  Find  the  angles  B  and  C 
and  the  side  A  C.  f  Angle  B  =  64°  14'. 

Ans.  \  Angle  C  =  79°  36'. 
I  Side  A  C—  64.1  ft. 


(54)     (a)    What    is    the    supplement    of    72C 
(b)  What  is  the  complement  of  22°  34'  17"  ? 


ir    36"? 

L07°  48r  24". 
67°  25'  43". 


(55)  In  a  triangle  ^  B  C,  angle  ^  =  57°  34. 5',  angle 
C  =  44°  22.5'  and  side  A  B  =  344  feet.  Find  angle  B  and 
the  sides  A  C  and  £  C.  f  Angle  ^  =  78°  3'. 

Ans.  -I  Side^  C"  =  481.22ft. 
<T  =  415.19ft 


PHYSICS. 


(1)  A  cubic  foot  of  a  certain  mineral  weighs  127  pounds; 
what  is  its  specific  gravity  ?  Ans.   2.034. 

(2)  A  conductor  conveying  a  current    of  electricity  is 
placed  in  a  horizontal  plane  pointing  north  and  south.     If 
the  north  pole  of  a  compass  needle  tends  to  point  towards 
the  east  when  the  compass  is  placed  directly  under  the  con- 
ductor, in  what  direction  is  the  current  flowing  in  the  con- 
ductor ? 

(3)  If  the  specific  gravity  of  bismuth  is  9.823,  what  is  the 
weight  of  a  cubic  inch  of  it  ?  Ans.   .3548  Ib. 

(4)  (a)  Give   a    short    description   of    a   saccharimeter. 
(b)  What  is  its  principal  use  ? 

(5)  (a)  What  is  science  ?     (b)  How  is  natural  science' dis- 
tinguished from  physical  science  ? 

(6)  What  chemical  actions  take  place  in  a  simple  voltaic 
cell  in  which  the  electrolyte  is  sulphuric  acid  diluted  with 
water  and  the  electrodes  are  made  of  copper  and  zinc  ? 

(7)  (a)  State    what    you    understand    by    polarization. 
(b)  What  is  meant  by  the  angle  of  polarization  ? 

(8)  Where  will  a  body  weigh  more,  on  the  top  of  a  high 
mountain  or  at  the  bottom  of  a  deep  valley,  the  bottom  of 
the  valley  being  as  far  below  sea  level  as  the  top  of  the 
mountain  is  above  ? 

§  4 


2  PHYSICS.  §  4 

(9)  The  air  contained  in  a  closed  vessel  under  a  pressure 
of  12  pounds  per  square  inch  is  heated  from  60°  to  300° ; 
what  is  its  final  tension?  Ans.   17.54  Ib. 

(10)  What   are   magnets,    and   from   what   is   the  name 
derived  ? 

(11)  (a)  Define  gravity,     (b)  Define  cohesion. 

(12)  A  charge  of  400  units  of  electricity  is  imparted  to  a 
sphere  4  inches  in  diameter;  what  is  the  electric  density  of 
the  charge  per  square  inch  over  the  surface  ? 

Ans.   7. 9577  units  of  electricity  per  square  inch. 

(13)  (a)  State  the  laws  of  reflection,     (b)  Would  these 
laws  hold  good  if  the  reflecting  surface,  instead  of  being  a 
plane,  were  curved — as  a  portion  of  a  sphere  ? 

(14)  A  certain  quantity  of  air  under  a  pressure  of  1^  atmos- 
pheres and  at  a  temperature   of  75°  weighs  7.14  pounds. 
It  expands  under  the  action  of  heat  until  1  cubic  foot  weighs 
.08  pound,      (a)  What  is  the  new  volume  ?     (b)  What  is  the 
new  temperature,   the  pressure  remaining    the  same  ?     (c) 
What  was  the  original  volume  ?  |"  (a)     89. 25  cu.  ft. 

Ans.    j  (£)     283.887°. 

I  (c)      64.188  cu.  ft. 

(15)  (a)  What  is  linear  expansion  ?     (b)  What  is  surface 
expansion  ?     (c)  What  is  cubical  expansion  ? 

(16)  If  2£  pounds  of  lead  having  a  temperature  of  40°  are 
placed  in  4  pounds  of  water  having  a  temperature  of  65° 
and  contained  in  a  cast-iron  vessel  whose  weight  is  If  pounds 
and  temperature  is  62°,  what  will  be  the  temperature  of  the 
mixture?  Ans.    64.43°. 

(17)  Define  specific  gravity. 

(18)  What  is  a  Leyden  jar  ? 

(19)  A  certain  volume  of  air  under  a  pressure  of  3£  atmos- 
pheres weighs  13  pounds.      After  expanding  at  a  constant 
temperature,   the  weight  of  an  equal  volume  is  only  2  pounds. 
What  is  the  final  pressure  of  the  air  ? 

Ans.   7.915  Ib.  per  sq.  in. 


§  4  PHYSICS.  3 

(20)  What  is  an  electrophorus  ? 

(21)  (a)  Of  what  is  a  mass  constituted  ?     (/>)   How  is  a 
molecule  constituted  ? 

(22)  (a)  Define  absolute  zero,     (b)  Define  absolute  tem- 
perature. 

(23)  A  compass  C  is  placed  between  the  north  and  south 
poles  of  two  magnets,  as 

shown  in  Fig.  1.  Towards 
which  pole  will  the  north 
pole  of  the  compass 
needle  tend  to  point,  and 
why  ?  FIG.  i. 

(24)  The  current  in  a   horizontal    conductor  is  flowing 
from  the  north  towards  the  south.      In  what  direction  will 
the  north  pole  of  a  compass  needle  point  if  the  compass  is 
placed  under  the  conductor  ? 

(25)  What  do  you  understand  by  a  liquid  body  ? 

(26)  If  a  body  weighs  1  pound  at  a  distance  100  miles 
from  the  center  of  the  earth,  what  will  it  weigh  (a)   at  the 
surface  ?  (b)  at  100  miles  above  the  surface  ?    Take  the  earth's 

radius  as  4,000  miles.  (  (a)     40  Ib. 

Ans.   '  v  ' 


(b)     38.0731b. 

(27)  Define  (a)  electrostatic;   (b)  electrodynamic. 

(28)  How  far  above  the  surface  of  the  earth  will  a  2-pound 
ball  weigh  3.  ounces  ?  Ans.   9,064  mi. 

(29)  Referring  to  electricity,   what  is  meant  by  a  con- 
tinuous current  ? 

(30)  If  4  cubic  feet  of  copper  alloy  weigh  2,000  pounds, 
(a)  what  is  the  specific  gravity  ?     (b)  What  is  the  weight  of 
a  cubic  inch  ?  .        (  (a)     Sp.  Gr.  =  8. 

*'\(b)     .28941b. 

(31)  Two  gilt  balls  A  and  B  are  statically  charged  with 
+  4  and— 24  units  of  electricity,  respectively;  if  the  balls 
are  allowed  to  touch  momentarily  and  are  then  placed  at  a 


4  PHYSICS.  §  4 

distance  of  2  centimeters  from  each  other,  will  they  attract 
or  repel  each  other,  and  what  will  be  the  force  in  dynes 
exerted  between  them  ? 

Ans.   Repelled  with  a  force  of  25  dynes. 

(32)  State  what  you  understand  by  inertia. 

(33)  Under  what  conditions  does  electricity  appear  ? 

(34)  A  piece  of  wood  weighs  11^  ounces  in  the  air.     It 
is  attached  to  a  piece  of  marble  weighing  5  pounds  in  the 
air  and  3  pounds  2  ounces  in  water.      Both  together  weigh 
2  pounds  9  ounces  in  water.     What  is  the  specific  gravity 
(a)  of  the  wood  ?  (b)  of  the  marble  ?  (  (a)     .555. 

AnS'|(£)      2.667. 

(35)  What  will  be  the  sign  of  the  static  charge  left  on  the 
cover  of  an  electrophorus,  if  the  cake  is  made  of  sulphur 
instead  of  resin  and  is  rubbed  with  a  piece  of  silk  ? 

(36)  A  bottle  weighs  2  pounds  in  air  and  10  ounces  in 
water.     A  pound  of  sugar  is  put  into  the  bottle,  and  the 
bottle  then  weighs  16  ounces  in  water.     What  is  the  specific 
gravity  of  the  sugar?  Ans.  '1.6. 

(37)  What  will  be  the  sign  of  the  static  charge  developed 
(a)  on  a  glass  rod  when  rubbed  with  fur  ?  (b)  on  a  piece  of 
hard  rubber  when  rubbed  with  silk  ?  (c)  on  a  piece  of  flannel 
when  rubbed  against  a  piece  of  amber  ? 

(38)  The  weight  necessary  to  sink  a  Nicholson's  hydrom- 
eter to  a  fixed  point  on  the  rod  is  2  pounds  8J  ounces.   The 
weight  necessary  to  sink  the  hydrometer  to  this  same  point 
when  a  piece  of  slate  is  in  the  basket  is  1  pound  11  ounces, 
and  when  the  slate  is  in  the  upper  pan,  12  ounces,    (a)  What  is 
the  specific  gravity  of  the  slate  ?     (b)  What  is  its  volume  ? 

(39)  For  what  purposes  is  an  electroscope  used  ? 

(40)  What  is  a  vacuum  ?     Illustrate  it. 

(41)  A  closed  vessel  fitted  with  a  piston  contains  oxygen 
under  a  pressure  of  3  atmospheres.   If  the  piston  is  so  moved 
that  the  volume  is  2J  times  its.  former  volume,  what  is  the 


§  4  PHYSICS.  5 

tension  of  the  gas  in  pounds  per  square  inch,  the  tempera- 
ture remaining  the  same  ?  Ans.   17.  G4  Ib.  per  sq.  in. 

(42)  For  what  is  the  torsion  balance  used  in  electrical 
measurements  ? 

(43)  If  you  are  told  that  the  vacuum  gauge  of  a  condenser 
shows  23  inches  vacuum,   what  do  you  understand  by  it  ? 
What,  in  that  case,  is  the  pressure  in  the  condenser  ? 

(44)  Two  gilt  balls  are  statically  charged  with  -f-  20  and 
—  5  units  of  electricity,  respectively ;  with  what  force  in  dynes 
will  they  attract  each  other  when  placed  (a)  at  a  distance  of  1 
centimeter  from  each  other  ?  (b)  at  a  distance  of  5  centimeters  ? 

Ans.  \  W     10°  dynes' 
(  (b)     4  dynes. 

(45)  What  is  the  weight  of  100  cubic  feet  of  hydrogen 
having  a  temperature  of  80°  and  a  tension  of  18  pounds  per 
square  inch?  Ans.   .6231  Ib. 

(46)  Four  cubic  feet  of  air  are  heated  under  a  constant 
pressure  from  40°  to   115°.     What  is  the  resulting  volume? 

Ans.  4.6  cu.  ft. 

(47)  From  what  is  the  word  electricity  derived  ? 

(48)  The   temperature   of    the  discharged  air  of  an  air 
compressor  is  120°  and  the  pressure  is  40  pounds  per  square 
inch  ;  when  it  has  cooled  to  the  temperature  of  the  surround- 
ing air,  which  is  55°,  what  is  its  pressure  ? 

Ans.   35.517  Ib.  per  sq.  in. 

(49)  State  the  principal  sources  of  heat. 

(50)  Describe  Nichol's  prism. 

(51)  What  is  the  weight  of  a  cubic  foot  of  air  at  60°  under 
a  pressure  of  1  atmosphere  ?  Ans.   .0763  Ib. 

(52)  State  the  law  of  attraction  and  repulsion  of  statically 
charged  bodies. 

(53)  State,  in  your  own  words,  what  you  understand  by 
the  "  temperature  of  a  body." 


6  PHYSICS.  §  4 

(54)  The  temperature   of  a  wrought-iron  bar  4  inches 
square  and  20  inches  long  is  raised  1,200°.     What  is  (a)  its 
cubical    expansion  ?     (b)    its    linear   expansion  ?      (c)    How 
much  larger  will  the  area  of  a  cross-section  be  ? 

((a)     7. 903  cu.  in. 
Ans.  I  (b)     .16464  in. 

I  (c)     .2634  sq.  in. 

(55)  What  do  you  understand  (a)  by  a  plano-convex  lens  ? 
(b)  by  a  concavo-convex  diverging  lens  ?     (c)  How  are  lenses 
constructed  ? 

(56)  WThat  are  the  physical  properties  of  electricity  ? 

(57)  Two  vessels,  the  volumes  of  which  are  each  7^  cubic 
feet,  are  filled  with  air;  the  temperature  is  the  same  in  both, 
but  the  tension  in  one  is  2  atmospheres  and  in  the  other 
40  pounds  per  square  inch.     If  all  the  air  in  one  vessel  is 
compressed  into  the  other,  what  is  the.  tension  of  the  mix- 
ture after  it  has  cooled  down  to  the  original  temperature  ? 

Ans.   69.4  Ib.  per  sq.  in. 

(58)  Three  gases — oxygen,  hydrogen,  and  nitrogen — are 
mixed  together  in  an  iron  retort  containing  40  cubic  feet. 
The  volume  and  the  tension  of  the  oxygen  are  12  cubic  feet 
and  1  atmosphere,  respectively;  of  the  hydrogen,  10  cubic 
feet  and  2  atmospheres;   of  the  nitrogen,  8  cubic  feet  and 
3  atmospheres.     The  temperature  of  the  separate  gases  and 
of  the  mixture  remaining  the  same  throughout,  what  is  the 
tension  of  the  mixture  ?  Ans.  20.58  Ib.  per  sq.  in. 

(59)  How  much  would  a  steel  wire  rope  900  feet  long 
shorten,  if  cooled  from  90°  to  28°  ?  Ans.  4.01  in. 

(60)  Define  magnetic  density. 

(61)  (a)  Name  the  kind  of  thermometer  in  general  use. 
Reduce  44°  R.  (b)  to  the  corresponding  centigrade  tempera- 
ture ;  (c)  to  the  corresponding  Fahrenheit  temperature. 

(  (b)     55°  C. 
AnS-)(,)     131°  F. 

(62)  Define  (a)  good  conductors;  (b)  bad  conductors. 


§  4  PHYSICS.  7 

(63)  State  what  is  meant  by  terrestrial  heat. 

(64)  Show  in  a  rough  sketch   (a)  the  refracting  edge, 
(b)  the  refracting  angle,  and  (c)  the  base,  of  a  prism. 

(65)  A  compass  C  is  placed  alongside  of  a  bar  magnet 
opposite  the  neutral  line,  as  shown  in  Fig.  2.    Towards  which 
pole  of  the  magnet  will  the  south 

pole  of  the  compass  needle  tend 
to  point,  and  why  ? 


(66)  A   vessel    containing    13 
cubic  feet  of  air,  having  a  tem- 
perature of  73°  and  a  tension  of  1  atmosphere,  is  placed  in 
communication  with  another  vessel  containing  18  cubic  feet 
of  air,  at  a  temperature  of  53°,  and  a  tension  of  30  pounds 
per  square  inch.     What  is  the  new  temperature,  if  the  ten- 
sion of  the  mixture  is  20  pounds  per  square  inch  ? 

Ans.   —20.65°. 

(67)  When  you  rub  a  piece  of  sulphur  with  a  piece  of 
flannel,  what  charge  does  the  sulphur  receive  ? 

(68)  Forty  cubic  feet  of  air  having  a  temperature  of  100° 
and  a  tension  of  90  pounds  per  square  inch,  are  mixed  with 
57  cubic  feet  of  air  having  a  temperature  of  130°  and  a 
tension  of  80  pounds  per  square  inch.     The  tension  of  the 
mixture  is  120  pounds  per  square  inch  and  the  temperature 
is  110°.     What  is  the  volume  ?  Ans.   67.248  cu.  ft. 

(69)  Change  (a)  -10°  R,  (b)  25°  F.,  and  (c)  2,200°  F.  into 
the  corresponding  centigrade  readings. 

Ha)     -23i°C. 
Ans.«(£)    -3f°C. 
[(c)     1,204|°C. 

(70)  Give  the  names  of  the  known  magnetic  substances. 

(71)  Twelve  cubic  feet  of  gas  are  heated  from  65°  to  390° ; 
what  is   the   increase   in  volume,  the   pressure   remaining 
constant?  Ans.   7.4286  cu.  ft. 

(72)  Define  (a)  electrical  resistance ;  (b)  conductivity. 


8  PHYSICS.  §  4 

(73)  (a)   Change  798    B.  T.  U.  to  calories,      (b)   Change 
40  calories  to  B.  T.  U.  j  (a)     201.515  calories. 

*'\(b)     158.4  B.T.U. 

(74)  A  piston  rod  is  4  inches  in  diameter  and  a  steel  piston 
is  bored  3.9985  inches  in  diameter  to  receive  it.     To  what 
temperature  must  the  piston  be  heated,  assuming  its  original 
temperature  to  be  80°  and  the  diameter   of  its  bore  after 
heating  to  be  4.001  inches,  to  allow  the  rod  to  enter  freely  ? 

Ans.   184.4°. 

(75)  If,  in  the  last  example,  the  whole  piston  were  raised 
to  the  same  temperature  as  the  bore,  and  its  weight  were  360 
pounds,  how  many  units  of  heat  would  be  required,  assuming 
a  loss  of  12$  from  various  causes  ?        Ans.   4,975.6  B.  T.  U. 

(76)  In  order  to  find  the  temperature  of  a  stove  fire,  a 
piece  of  cast  iron  weighing  1  pound  is  placed  in  it.     A  copper 
vessel  weighing  1|  pounds  is  partly  filled  with  3^  pounds  of 
water,  the  temperature  of  both  being  85°.     After  the  cast 
iron  piece  has  been  placed  in  the  vessel  of  water,  the  tem- 
perature of  the  mixture  is  found  to  be  128°.     What  was  the 
temperature  of  the  fire  ?  Ans.   1,252°. 

(77)  A  solution  of  cane  sugar  placed  in  a  tube  12  inches 
long,  and  as  observed  through  the  saccharimeter,  requires 
an  angle  of  rotation  of  29°  to  reproduce  entire  darkness. 
What  per  cent,  of  sugar  is  contained  in  this  solution  ? 

Ans.   12.33$. 

(78)  What  is  supposed  to  be  the  cause  of  all  electrical 
phenomena  ? 

(79)  What  do  you  understand  by  "specific  heat  "  ? 

(80)  A  vessel  has  a  volume  of  6.7  cubic  feet.     A  vacuum 
gauge  attached  to  it  shows  17J  inches.     How  much  air  at 
atmospheric  pressure  will  it  be  necessary  to  admit  to  have 
the  vacuum  gauge  show  (a)  5  inches  ?  (b)  0  inches  ? 

((a)     2.791  cu.  ft. 
"  {  (b)     3.908  cu.  ft. 

(81)  If  4,516  cubic  inches  of  gas  having  a  temperature  of 


§  4  PHYSICS.  9 

260°  are  cooled  down  to  a  temperature  of  80°,  the  pressure 
remaining  the  same,  what  is  its  new  volume  ? 

Ans.   1.06  cu.  ft. 

(82)  (a)  What  do  you  understand  by  dispersion  of  light  ? 
(b)  Into  what  colors  is  light  decomposed  ?     (c)  What  infer- 
ence has  been  drawn  from  the  presence  of  Fraunhofer's  lines 
in  the  solar  spectrum  ? 

(83)  What  is  sensible  heat  ? 

(84)  How  many  units  of  heat  would  be  required  to  vapor- 
ize  12  pounds  of  solid  mercury  having  a  temperature  of 
—  50°  ?     (Use  the  same  value  for  the  specific  heat  of  solid 
mercury  as  given  in  the  table  for  liquid  mercury.) 

Ans.   2,229.595  B.  T.  U. 

(85)  What  is  the  absolute  temperature  corresponding  to 

(a)  96°  F.  ?   (b)  32°  C.  ?  (c)  180°  C.  ?  (d)  650°  F.  ?  (e)  -  40°  C.  ? 

(86)  (a)  Name  the  rays  produced  by  double  refraction. 

(b)  Name  a  mineral  which  possesses  the  property  of  double 
refraction. 

(87)  What  is  meant  by  the  term  quantity  of  heat  ? 

(88)  State  what  you  know  of  the  theory  of  light. 

(89)  State  what  you  know  about  a  magnetic  field. 

(90)  (a)  What  do  you  understand  .by  a  translucent  body  ? 
(b)  What  is  a  homogeneous  medium  ? 

(91)  (a)  What  is  meant  by  polarization  in  opposite  direc- 
tions ?     (b)  Which  media  possess  the  property  of  polarization  ? 

(92)  How  can  you  explain  the  quivering  of  objects  seen 
over  very  hot  bodies  ? 

(93)  (a)  What  do  you  understand  by  circular  polarization  ? 
What  does  it  mean  if  a  substance  is  said  to  be  levo-rotary  ? 

(94)  What  is  the  approximate  velocity  of  light  ? 

(95)  What  do  you  understand  by  the  index  of  refraction  ? 

(96)  State  some  experiments  for  detecting  the  presence 
of  static  charges  of  electricity. 


THEORETICAL  CHEMISTRY. 


(1)  Calcium   carbonate  CaCO3  may  be   precipitated   by 
heat,  the  reaction  being  represented  by  the  following  equa- 
tion : 

CaHJ^CO^    =     CaCO,    +    Hft    +     CO, 

calcium  calcium  carbon 

bicarbonate  carbonate  dioxide 

'How  much  calcium  carbonate  would  be  precipitated  by 
heating  296.3  grams  of  calcium  bicarbonate  ? 

Ans.   182.9  grams. 

(2)  Explain  what  you  understand  by  chemical  combina- 
tion in  definite  and  multiple  proportion. 

(3)  Give  a  short  account  of  how  you  would  proceed  to 
weigh  a  test  tube  on  a  chemical  balance. 

(4)  Define  the  terms  inflammable  body  and  supporter  of 
combustion  ;  explain  why  these  terms  are  purely  relative. 

(5)  How  many  atoms  are  contained  (a)  in  a  molecule  of 
mercury  ?  (b)  in  a  molecule  of  ozone  ?  (c)  in  a  molecule  of 
arsenic  ? 

(6)  A  white  powder  is  shaken  up  with  distilled  water; 
how  would  you  ascertain  whether  any  of  it  is  dissolved  ? 

(7)  State  Avogadro's  law  and  the  deductions  therefrom. 

(8)  Define  meta-acids. 

(9)  What  is  meant  by  the  term  allotropy  ? 


2  THEORETICAL  CHEMISTRY.  §  5 

(10)  The  analysis  of  a  compound  of  chlorine  and  antimony 
gives  46. 61$  of  chlorine  and  53. 39$  of  antimony;  the  density 
of  this  compound,  is  114.25.     What  is  the  atomic  weight  of 
antimony  ? 

(11)  Define  matter.     Describe  its  general  properties  and 
various  physical  states. 

(12)  Define  combustion,  and  state  what  you  understand 
by  that  term. 

(13)  Explain  the  meaning  of  the  terms  analysis  and  syn- 
tJicsis. 

(14)  In  which  way  does  methyl-orange  differ  from  phenol- 
phthalein  ? 

(15)  Which  of  the  following  elements  are  metalloids  and 
which  are  metals:   Cl,  K,  P,  S,  N,  Na,  As,  H,  Fe  ? 

(16)  (a]  What  do  you  understand  by  a  compound  radical? 
(b)  How  are  compound  radicals  named  ? 

(17)  Give  examples  of  oxidizing  and  reducing  agents. 

(18)  Define  force,  and  state  why  chemical  action  may  be 
properly  regarded  as  a  force. 

(19)  What  is  molecular  volume,  and  how  is  it  expressed  ? 

(20)  What  is  the  percentage  composition  of  chromic  acid 
H^CrO^  ?  f     1. 69$  of  hydrogen. 

Ans.  -j  44.30$  of  chromium. 
I  54.01$  of  oxygen. 

(21)  A  certain  volume  of  oxygen,   under  a  barometric 
pressure  of  781  millimeters,  measures  542  cubic  centimeters. 
What  is  its  true  volume,  measured  at  760  millimeters  ? 

Ans.   556.97  c.  c. 

(22)  What  is  chemistry? 

(23)  A  glassful  of  dirty  water  is  given  to  you.      Describe 
the  plan  you  would  adopt,  first,  for  rendering  the  water  clear, 
and,    secondly,    for  ascertaining  whether  or  not  the  clear 
water  contains  any  dissolved  matter. 


§  5  THEORETICAL  CHEMISTRY.  3 

(24)  State  the  difference  between  a  mechanical  mixture 
and  a  chemical  combination. 

(25)  Mention  the  elements  that  at  ordinary  temperatures 
are,  respectively,  gases  and  liquids. 

(26)  State   what   you   know   of   chemical   and    physical 
changes. 

(27)  How  would  you  prepare  salt  from  sea-water  ? 

(28)  An  analysis  of  a  body  having  a  molecular  weight  of 
84  gave  the  following  results : 

C=  14.290 

O  =  5  7.1  40 

H=  1.190 

Na  =  2  7.3  80 

1  00.000 
What  is  its  chemical  formula  ?  Ans.   HNaCO^. 

(29)  State  what  you  know  about  atomicity. 

(30)  What  relation,   if  any,    exists  between   the  atomic 
weight  and  the  specific  heat  of  bodies  ? 

(31)  (a)  Define  the  terms  precipitation,  precipitate,  and 
precipitant,      (b)  Give  Berthollet's   law  concerning  precip- 
itation. 

(32)  How  many  volumes  of  carbon  dioxide   CO^  will  be 
formed  when  a  mixture  of  4  volumes  of  carbon  monoxide 
CO  and  4  volumes  of  oxygen  O  is  burned  ?     Show  the  reac- 
tion through  an  equation ;  state  what  volume  of  oxygen,  if 
any,  remains  unburned. 

(33)  Explain  the  meaning  of  the  terms  sublimation,  vol- 
atile, fixed,  and  ignition. 

(34)  What  are  the  effects  on  matter  of  gravitation,  cohe- 
sion, and  chemical  attraction,  respectively  ? 

(35)  Define  the  terms  acid,  base,  and  salt. 

(36)  Explain  the  meaning  of  the  term  nascent. 


4  THEORETICAL  CHEMISTRY.  §  5 

(37)  Give  a  short  account  of  the  metric  system  of  weights 
and  measures. 

(38)  By  what  properties  is  a  metal  distinguished  from  a 
non-metal  ? 

(39)  Define  the  terms  monobasic  and  dibasic  as  applied  to 
acids. 

(40)  What    is    the    percentage     composition     of     borax 
NatB4  O,  ?  f  22. 77$  of  sodium. 

Ans.  j  21.78$  of  boron. 
155.45$  of  oxygen. 

(41)  (a)  What   is    expressed   by   a    chemical   equation  ? 
(b)  How  is  such  an  equation  constructed  ? 

(42)  What  do  you  understand  by  a  fluid  ounce  ? 

(43)  State  what  you  know  about  the  nomenclature   of 
acids  and  salts. 

(44)  What  are  stoichiometrical  calculations  ? 

(45)  Define  density. 

(46)  (a)  How   many  atoms  may  a  compound  molecule 
contain  ?     (b)  How  are  compound  molecules  classified  ? 

(47)  (a)  State    Gay-Lussac's   two   laws    concerning    the 
proportions  in  which  gaseous  volumes  combine,      (b)  How 
may  these  laws  be  derived  from  Avogadro's  law  ? 

(48)  (a)  In    what   way   may    chemical    changes    occur  ? 
(b)  Illustrate  these  changes  by  using  the  letters  A,  B,   etc. 
instead  of  symbols. 

(49)  Ammonium  iodide  may  be  prepared  according  to 
the  equation : 

NH       +       HI      =       NH I 

3  4 

ammonia  hy^adlC         ^SuSe"111 


If  56. 1  grams  of  ammonia  yield  478. 5  grams  of  ammonium 
iodide,  and  the  molecular  weight  of  ammonium  iodide  is 
145,  what  is  the  molecular  weight  of  ammonia  ?  Ans.  17. 


§  5  THEORETICAL  CHEMISTRY.  5 

(50)  Explain  the  terms  crystal,  dimorphous,  amorphous, 
water  of  crystallization,  and  isomorphous. 

(51)  (a)  Define  molecule,     (b)  State  how  molecules  are 
classified. 

(52)  What  do  you  understand  (a)  by  a  symbol  ?  (b)  by  a 
formula  ? 

(53)  (a)  How  do  atoms  of  different  valence  combine  ? 
(b)  Suppose  a  pentad  and  a  dyad  combine,  how  many  atoms 
of  each  are  necessary  to  form  a  saturated  compound  ? 

(54)  In   what   respects   does   a   mixture   of  copper  and 
sulphur  differ  after  being"  heated  from  its  condition  previous 
to  the  application  of  heat  ? 

(55)  A  compound,  hydrodisodium  phosphate,  is  found  on 
analysis  to  contain 

hydrogen  =  0.  7  % 

sodium  =  3  2.4$ 

phosphorus  =  2  1.8$ 

oxygen  =  4  5.1  % 

1  00.0$ 
Its  molecular  weight  is  142.     What  is  its  formula  ? 

Ans. 


(56)  What  is  the    simplest  way  to  distinguish  between 
elemental  and  compound  bodies  ? 

(57)  Why   does  chemical   combination  always  occur  in 
either  definite  or  multiple  proportion  ? 

(58)  Alumina  is  composed  as  follows  : 

aluminum  =    5  3.4$ 
oxygen  =    4  G.  6  $ 
1  00.0$ 
Its  molecular  weight  is  103.     What  is  its  formula  ? 

Ans.   Al9Of 

(59)  Potassium  sulphate  may  be  obtained  according  to 
the  following  equation  : 


THEORETICAL  CHEMISTRY. 


2KNO,      + 

potassium  sulphuric  potassium  nitrous 

nitrite  acid  sulphate  acid 

How   much   potassium   nitrite  will  be  required  to  yield 
392  grams  of  potassium  sulphate  ?  Ans.   383  grams. 

(60)  Define  (a)  a  monad  ;    (b}  a  heptad  ;  (c)  a  pentad. 

(61)  Give  the  names  of  SrO,  NaCl,  P2<95,  £2<93,  SnCl^ 

(62)  A  molecule  of  potassium  nitrate  contains  1  atom  of 
potassium,  which  is  by  weight  38.  7$  of  it.     The  molecular 
weight  of  potassium   nitrate  is   101.     What  is  the  atomic 
weight  of  potassium  ?  Ans.   39. 

(63)  Describe  the  process  of  gaseous  diffusion,  and  state 
the  law  governing  its  rate. 

(64)  What   do    you    understand   by   an   electronegative 
element  ? 

(65)  Define  the  meaning  of  the  term  perissad^ 

(66)  State  how  acids  are  classified. 

(67)  A  molecule  of  zinc  carbonate  contains  1  atom  of 
zinc,  which  is  by  weight  52$  of  it  ;  the  molecular  weight  of 
zinc  carbonate  is  125.     What  is  the  atomic  weight  of  zinc  ? 

Ans.   65. 

(68)  What  will  a  gas  measure  at  0°,  which  at  100°  measured 
40.  1  cubic  centimeters  ?  Ans.   29.  349  c.  c. 

(69)  State    how  you    would    symbolically    express    the 
valence  of  an  atom. 

(70)  (a)  State  the  principles  that  underlie  the  nomencla- 
ture of  binary  molecules,     (b}  Which  termination  is  always 
characteristic  of  a  binary  molecule  ? 

(71)  Why  may  mass  reactions  be  represented  by  molecular 
formulas  ? 

(72)  Zinc  sulphide  ZnS  contains  67$  of  zinc,  or  1  atom  of 
zinc  in  each  molecule.     The  atomic  weight  of  zinc  is  65  ; 
what  is  the  molecular  weight  of  zinc  sulphide  ?         Ans.   97. 


§  5  THEORETICAL  CHEMISTRY.  7 

(73)  Phosphonium  iodide  PHJ  may  be  obtained  accord- 
ing to  the  following  equation : 

PH3      +      HI      =       PHJ 

hydrogen         hydriodic        phosphonium 
phosphide  acid  iodide 

If  883.2  grams  of  hydriodic  acid  yield  1,117.8  grams  of 
phosphonium  iodide,  whose  molecular  weight  is  162,  what  is 
the  molecular  weight  of  hydriodic  acid  ?  Ans.  128. 

(74)  What  volume  would  350  cubic  centimeters  of  ammonia 
gas,  measured  at  74°,  have  at  94°  ?  Ans.   370.17  c.  c. 

(75)  Define  atomic  weight. 

(76)  Write  in  words  the  meaning  of  each  of  the  following 
symbols:  H,  O,  6>2,  3Hf 

(77)  How  many  liters  of  hydrogen  weigh  1  gram  ? 

(78)  Ammonium  nitrate  NH^NO^  decomposes  under  the 
influence  of  heat  into  one  molecule  of  hyponitrous  oxide  NtO 
and  2  molecules  of  water  H^O.     How  many  parts  of  hyponi- 
trous oxide  N^O  are  there  in  100  parts  of  ammonium  nitrate  ? 

Ans.   55  parts. 

(79)  Antimonous  sulphide  may  be  precipitated  according 
to  the  following  equation  : 

KSb'Cl^    +    3Sfft    =      KHCl       +       Sb^ 

antimonous         hydrogen         hydrochloric         antimonous 
chloride  sulphide  acid  sulphide 

78-57  grams  of  antimonous  sulphide  are  obtained  from  105. 6 
grams  of  antimonous  chloride,  the  molecular  weight  of  which  is 
228. 5.  What  is  the  molecular  weight  of  antimonous  sulphide  ? 

Ans.   340. 

(80)  You  are  required  to  neutralize  with  caustic  soda  a 
quantity  of  sulphuric  acid  diluted  with  water.     Describe  how 
you  would  do  this,  and  how  you  would  ascertain  that  the 
neutralization  has  been  effected. 

(81)  State  what  the  valence  of  an  atom  means. 

(82)  What  data  are  necessary  to  determine   the  atomic 
weight  of  an  element  ? 


8  THEORETICAL  CHEMISTRY.  §  5 

(83)  Salt  NaCl  contains  39.32$  of   sodium,   the  atomic 
weight  of  which  is  23  ;  what  is  the  molecular  weight  of  salt  ? 

Ans.   58.5. 

(84)  Sodium  sulphate  may  be  formed  according  to  the 
following  equation: 

ff,S04  +  NafO,  =  Na^SO,  +   CO,   +    H^O 

sulphuric  sodium  sodium  carbon 

acid  carbonate        sulphate        dioxide 
98  106  142  44  18 

How  much  sodium  sulphate  may  be  yielded  by  the  decom- 
position of  396.11  grams  of  sodium  carbonate  ? 

Ans.   530. 64  grams. 

(85)  (a)  What  do  you  understand  by  the  density  of  a  gas  ? 
(b)  The  volume  of  a  given  weight  of  a  gas  has  to  be  calcu- 
lated at  the  standard  of  a  temperature  and  pressure ;  what 
does  this  mean  ? 

(86)  What  is  (a)   a  chemical    reaction  ?    (b)  a  chemical 
reagent  ? 

(87)  How  much  calcium  phosphate  is  required  to  yield 
501  kilograms  of  phosphorus  ?     The  atomic  weight  of  phos- 
phorus is  31,  and  the  molecular  weight  of  calcium  phosphate 
is  310.  Ans.   5,010  Kg. 

(88)  What  volume  would  420  cubic  centimeters  of  hydro- 
gen, measured  at  739  millimeters  pressure,   occupy  at  760 
millimeters  pressure  ?  Ans.   408.4  c.  c. 

(89)  (a)  What  are  the  properties  common  to  acids  ?  (b)  In 
what  respect  does  sulphuric  acid  differ  from  sulphurous  acid  ? 

(90)  (a)  How  are  chemical  reactions  classified  ?    (b)  Illus- 
trate by  using  the  letters  A,  B,  etc.  instead  of  symbols. 

(91)  How  much  does  1  liter  of  hydrogen  weigh  ? 

(92)  What  do  you  understand  by  molecular  stability  ? 


§  5  THEORETICAL  CHEMISTRY.  9 

(93)  Give  the  formulas  of  potassium  iodide,  lead  sulphide, 
calcium  chloride,  and  cupric  oxide. 

(94)  Read  the  following  equation  by  weight  : 


potassium     .       sulphuric         potassium  nitric 

nitrate  acid  sulphate  acid 

(95)  How  much  lead  maybe  obtained  from  932  kilograms 
of  lead  sulphide  ?  The  atomic  weight  of  lead  is  207  and  the 
molecular  weight  of  lead  sulphide  is  239.  Ans.  807.21  Kg. 


INDEX. 


NOTE.— All  items  in  this  index  refer  first  to  the  section  (see  Preface,  Vol.  I)  and  then 
to  the  page  of  the  section.  Thus,  "Combustion  5  86"  means  that  combustion  will  be 
found  on  page  86  of  section  5. 


A. 


Sec. 


Absolute  pressure  ................  4 

"          temperature  ...........  4 

Abstract  numbers  ...............  1 

Acid,  Definition  of  ................  5 

"       Normal,  and  basic  salts...  5 
"       sulphuric,    Determination 

of  ...........  ...............  5 

Acidimetry  and  alkalimetry  ----  5 

Acids  and  salts,  Nomenclature  of  5 

"     Basicity  of  ..................  5 

"     Classification  of  ............  5 

Action,  chemical,  Modes  of  ......  5 

Addition  ..........................  1 

"        in  algebra  ...............  3 

"         "decimals  .............  1 

"        of  denominate  numbers  2 

"         "fractions  ..............  1 

"         "          "         in  algebra..  3 

"         "  like  quantities  .......  3 

"         "monomials  ...........  3 

"         "polynomials  ..........  3 

Rule  for  .................  1 

"         Sign  of...-  ...............  1 

table  ................  ....  1 

Agents,  Oxidizing  ................  5 

"        Reducing  ................  5 

Aggregation,  Symbols  of  ........  1 

"  ........  3 

Air,  Properties  of  ................  4 

"    pumps  ........................  4 

"    unit,   relation   to  hydrogen 

unit  ........................  5 

Algebra,  Definition  of  ............  3 

"         Notation  used  in  .......  3 

"          Signs  used  in  ...........  3 

"          Use  of  letters  in  ........  3 

Algebraic  expressions,  Reading  3 


Page.  Sec.  Page. 

27  Algebraic  expressions, Terms  of  3  5 

31        Alkalies  and  bases 5  41 

1        Allotropy 5  81 

40        Amount  (Percentage) 2  2 

45        Analysis  and  synthesis 5  7 

Aneroid  barometer 4  23 

94        Angle,  Complement  of 3  62 

91             "       Functions  of 3  63 

43  "       Supplement  of 3  62 

44  "       To  find,  from  function...  3  64 

45  Angles  or  arcs,  Measurement  of  2  11 
59        Antecedent  of  a  ratio 2  43 

4        Apparatus,  List  of 5  97 

10        Arabic  notation 1  2 

38  Arithmetic,  Definition  of 1  1 

15  Fundamental    proc- 

27                                  esses  of 1  4 

39  Atom,  Definition  of 4  2 

12            "                "           " 5  3 

12        Atomic  group,  Calculation  of 5  67 

15              "       weight,  Definition  of 5  21 

8  "             "         Determination 

4  of 5  32 

5  "         Relation    of,    to 

89  specific  heat...  5  34 

90  "       weights,  Table  of 5  28 

49        Atomicity '...  5  20 

17  "          Table  of - 5  21 

18  Atoms  and  molecules 5  18 

37  "         "            "          Nomencla- 
ture of..  5  19 

71             "       Motion  of 4  3 

3  Attraction,  gravitation,  and  co- 

3  hesion,    Relation 

4  between 5  8 

1                 "            of  matter 4  2 

6  Avoirdupois  weight 2  9 

ix 


INDEX. 


B.  Sec.  Page. 
Balance,  The 5  13 

"        Torsion 4  105 

Barometer,  Aneroid 4  23 

Barometers 4  22 

Base  (Percentage) 2  2 

Bases  and  alkalies 5  41 

Basic,  normal,  and  acid  salts 5  45 

Basicity  of  acids 5  44 

Battery,  Electrostatic 4  115 

Voltaic 4  118 

Berthollet's  laws 5  58 

Binary  compounds,  Formation  of  5  38 
"            Nomencla- 
ture of 5  36 

Binomial,  Definition  of 3  6 

Bodies,  Solid,  liquid,  and  gaseous  4  3 

Brace 1  49 

Brackets 1  49 

British  thermal  unit 4  58 

Brittleness,  Definition  of 4  6 

Bunsen  burner 5  99 

Burettes 5  93 

"         Practical      suggestions 

concerning 5  108 

C.  Sec.  Page. 
Calculation  of  atomic  group 5  67 

"  percentage    com- 
position    5  64 

Calculations  from  equations 5  68 

Calorie,  The 4.  58 

Cancelation 1  19 

"           inequations 3  53 

Rule  for 1  21 

Capacity,  Measures  of 2  10 

Cause  and  effect,  Principle  of...  2  54 

Cell,  Voltaic  or  galvanic 4  118 

Chain,  Engineer's. . 2  9 

"      Gunter's.... 2  8 

Charges,  Positive  and  negative  4  109 

Chemical  action,  Modes  of 5  59 

"             "       Relation  to  force  5  7 
"         and  physical  changes  5  4 
"        proper- 
ties. ...  5  5 

"        operations.... 5  81 

Chemistry,  Object  of 5  4 

Cipher 1  2 

Circular  polarization 4  98 

Clearing  equations  of  fractions  3  54 

Coefficient,  Definition  of 3  4 

"            of  expansion 4  49 

Cohesion,  attraction,  and  gravi- 
tation. Relation  between 5  8 

Combination  by  volume 5  50 

** .  Chemical,  and 
mechanical  mix- 
ture   5  5 


Sec.  Page. 

Combination  in  definite  propor- 
tions   5  25 

"            in  multiple  propor- 
tions   5  25 

Combining  power,  Quantity  of. .  5  21 

Combustion 5  86 

Supporters  of 5  87 

Common  denominator 1  26 

Compass,  The 4  97 

Complement  of  an  angle 3  62 

Complex  fractions 3  45 

Composition,  percentage,  Calcu- 
lation of 5  64 

Compound  denominate  number  2  8 

proportion 2  55 

Rule  for..  2  56 
"           radicals, Names  of...  5  40 
Compounds,  binary,  Nomencla- 
ture of 5  36 

"  Table  of  solubility 

of 5  60 

"             Ternary 5  40 

Compressibility,  Definition  of . . .  4  5 

Concrete  numbers 1  1 

Condenser,  Electrical 4  114 

Liebig's 5  84 

Conductivity,  Electrical 4  108 

Conductors,  Electrical 4  108 

Consequent  of  a  ratio 2  43 

Cosines  and  sines,  natural,  Table 

of.- 3  77 

Cosine  of  an  angle 3  63 

Cotangent  of  an  angle 3  63 

Cotangents   and  tangents,  nat- 
ural, Table  of 3  85 

Couplet  of  a  proportion 2  47 

"       (Ratio) 2  43 

Crystallization 5  84 

Crystallography 5  74 

Crystals,  Classification  of 5  79  . 

"         Systems  of. 5  79 

Cube  of  a  number 2  23 

!•      root 2  25 

"     2  32 

"    Proofof 2  38 

"    Rule  for 2  38 

Cubic  measure 2  9 

Currents,  electric,  Production  of  4  116 

D.                    Sec.  Page. 
Dates,  To  find  interval  of  time 

between 2  18 

Decantation 5  82 

Decimal  point 1  36 

"  To  express  approxi- 
mately as  a  fraction 
having  a  given  denom- 
inator   1  48 


INDEX. 


XI 


Decimal,  To  reduce  a  fraction  to 


Sec.  Page. 


To  reduce,  to  a  fraction 

Decimals 

Addition  of 

Division  of 

how  read 

Multiplication  of.... 1 

Subtraction  of 1  39 

Reduction  of 1  46 

Denominate  number,  Compound  2  8 

numbers 2  7 

"        Addition  of  2  15 
44        Division  of  2  .19 
"                   44        Multiplica- 
tion of....  2  19 
44                   "       Reduction 

of 2  12 

44        Simple 2  7 

Subtrac- 
tion of....  2  17 
44                   "       To  reduce, 
to  higher 
denomi- 
nations... 2  13 
44                   "       To  reduce, 
to     lower 
denomina- 
tions    2  12 

Denominator 1  22 

Density  and  molecular  weight..  5  47 

44        Definition  of 5  20 

41       Magnetic 4  101 

Destructive  distillation 5  87 

Determination    of    sodium    car- 
bonate   5  95 

44               of  sulphuric  acid  5  94 

Difference 1  9 

(Percentage) 2  2 

Diffusion  of  gases 5  49 

Digits 1  2 

Direct  proportion..^ 2  47 

•        ratio 2  43 

Directions  for  laboratory 5  96 

Dispersion  of  light 4  83 

Distillation 5  84 

"          Destructive 5  87 

Dividend 1  16 

Divisibility,  Definition  of 4  4 

Division 1  16 

*4        in  algebra. 3  25 

of  decimals. 1  42 

**         "  denominate  numbers  2  20 

•*         "fractions 1  32 

«                                3  42 

«*         "monomials 3  25 

"         "polynomials. 3  » 

Rulefor 1  1» 


Sec.  Page. 
Division,  Sign  of  .................      i       16 

Divisor  ............................      i        je 

Dollars  and  cents,  how  written 


49 


decimally 

44       Sign  for 

Double  refraction 

Dry  measure 

Ductility,  Definition  of 

Dynamical  theory  of  heat 

E.  Sec.  Page. 

Elasticity,  Definition  of 4         5 

Electric  currents,  Production  of  4      116 

44        series 4      104 

Electrical  conductivity 108 

44         conductors 108 

44         non-conductors 108 

44         potential 115 

44         resistance 108 

Electricity 101 

Electrochemical     character     of 

elements,  Table  of 31 

Electrodes  or  poles 118 

Electrodj-namics 115 

Electromagnet,  The 127 

Electromagnetism 122 

Electromotive  series 119 

Electrophorus,  The 110 

Electropositive  and  electronega- 
tive elements 30 

Electroscope,  The 104 

Electrostatic  battery 115 

44             machines 4      112 

Electrostatics 4      102 

Elemental  molecules,  Number  of  5        18 
Elements,    Electropositive    and 

electronegative 5       30 

Meta- 5        30 

44          Nascent  condition  of  5       87 

44          Table  of 5       28 

"          Table  of   electro- 
chemical character 

of 5       31 

Engineers'  chain 2         9 

Equality,  Sign  of 1         4 

Equations,  Calculations  from;..  5       68 

44          Cancelation  in 3       53 

Changing  signs  in ...  3       53 

Chemical 5       56 

Clearing  of  fractions  3       54 

Definition  of 3         3 

Literal 3       56 

Membersof 3       51 

Simple 3       55 

Transformations  in..  3       52 
"          with  one     unknown 

quantity 3       56 

Evaporation 5       82 


Xll 


INDEX. 


Sec.  Page. 


Evolution 2 

Expansibility,  Definition  of 4 

Expansion,  Coefficient  of 4 

of  bodies  by  heat 4 

44  gases 4 

Explosion 5 

Exponent 2 

44          3 

44         Fractional 3 

44         Negative 3 

Exponents,  Literal 3 

14  Theory  of 3 

Expressions,  algebraic,  Reading 

of 3 

Extension,  Definition  of 4 

44          Measures  of 2 

Extremes  (Proportion) 2 


F.  Sec. 

Factor,  Prime . 1 

Factoring 3 

Factors ...  1 


44       Equal 3 

Fathom 2 

Figures 1 

44       Local  or  relative  values 

of 1 

44       Simple  value  of 1 

Filtration 5 

Force 5 

44     Magnetic 4 

44     of  gravity 4 

44     Relation  of  chemical  action 

to 5 

Formulas  and  symbols 5 

44         for  gravity  problems  4 

Foucault's  prism 4 

Fraction 1 

44         Improper 1 

44        Lowest  terms  of 1 

44        Proper 1 

44         Signs  of 3 

44         Terms  of 1 

44        To  invert  a 1 

44         To  reduce  a  decimal  to  1 

44         To  reduce,  to  a  decimal  1 
44         To  reduce,  to  a  higher 

term 1 

44         To  reduce,  to  an  equal 
fraction  with  a  given 

denominator 1 

"        To     reduce,   to    lower 

terms 1 

44         Value  of 1 

Fractional  exponent 3 

Fractions,  Addition  of 1 


Page. 
20 
28 
19 

4 

29 
11 

2 

2 
2 


98 


34 


Fractions,  Addition  and  subtrac- 
tion of 

44           Clearing  of,  in  equa- 
tions  

44  Complex 

14  Division  of. . . 


Sec.  Page. 


in  algebra 

Multiplication  of, 


44          Reduction  of 1 

44                   "           44  3 

44           Roots  of 2 

44          Subtraction  of 1 

To  reduce,  to  common 

denominator 1 

Function,  To  find  angle  from 3 

Functions,  To  find,  from  angle. .  3 

Trigonometric 3 

Fusion  and  vaporization,  Table 

of  temperature  of 4 

Latent  heat  of 4 

G.  Sec. 

Gain  or  loss  per  cent 2 

Gallon,  C  ubic  inches  in 2 

44       Weight  of 2 

Gallons  in  cubic  foot 2 

Galvanic  or  voltaic  cell,  The 4 

Gaseous  bodies 5 

Gases,  Diffusion  of 5 

44       Expansion  of 4 

44       Mixtures  of. . .  4 


Permanent 


and 


Pressure,     volume, 

temperature  of 5 

44       Properties  of 4 

Specific  heat  of 4 

44       Table  of  specific  gravity 
of 

44       Tension  of 

Gay-Lussac's  laws 


Glass 

44     Directions  for  bending 

44  "    cutting 

44                           u   drawingout 
Gravitation,    cohesion,   and    at- 
traction,      Rela- 
tion between 5 

14  Law  of 4 

Gravity,  Force  of 4 

44        problems,  Formulas  for      4 

44         Specific 4 

44  of  gases,  De- 
termination 
of...  4 


65 

Page. 

6 

11 

11 

11 

118 

2 
49 
25 
35 


72 

18 
61 

11 
18 
30 
50 
99 
101 
100 
102 


Gravity,  specific,  of  liquids,  De- 
termination 

of 

44  44         44    solids,     De- 

termination 

of 

44  44         Tables  of 

Group,  atomic,  Calculation  of... 
Gunter's  chain 2 

H. 

Hardness,  Definition  of 4 

Heat 

"     conduction       by      metals, 

Table  of 

41     Dynamical  theory  of 4 

41      Expansion  of  bodies  by 

44     Latent '. 

41  44      of  fusion 4 

44        44  vaporization 

44     Measurement  of 4 

44     Nature  of 

44     Propagation  of 

"      Quantity  of 

44     Sources  of 

44     Specific 

44        of  gases,  Table  of.. 

44  4l         44  liquids,  Table  of 

44  l4        44  solids,  Table   of 

44  44        Relation       of,       to 

atomic  weight... 

Hydrogen,  Absolute  weight  of.. 

44  unit,  Relation  to  air 

unit 

Hydrometers 4 

I. 

Ignition 5 

Impenetrability,  Definition  of... 
Improper  fractions,  To   reduce, 

to  mixed  numbers 1 

Inches,   To  reduce,    to    decimal 

parts  of  a  foot 1 

Indestructibility,  Definition  of. . 

44  of  matter 

Index  of  a  root 

Inertia,  Definition  of 

Inflammable  bodies 

Insulators,  List  of 4 

Integer 1 

Integral  expression 3 

Inverse  proportion 2 

2 

44       ratio 2 

Involution 2 


K. 


Known  quantity . 


INDEX. 

Kill 

Sec. 

Page. 

L,. 

Sec. 

Page. 

Laboratory  directions  

5 

96 

Latent  heat  

4 

64 

4 

12 

44    of  fusion  

4 

65 

1     44  vaporization  

4 

66 

Law,  Mariotte's  

4 

26 

4 

13 

44               44         

5 

72 

4 

10 

44     of  gravitation  

4 

7 

5 

67 

44      44  refraction  

4 

76 

2 

8 

Laws,  Berthollet's  

5 

58 

44      Gay-Lussac's  

4 

30 

Sec. 

Page. 

44 

5 

50 

4 

5 

44       of  electrostatics  

4 

104 

4 

41 

44       44  reflection  

4 

74 

44       44  weight  

4 

7 

4 

56 

League  

2 

11 

4 

56 

Least  common  denominator,  To 

4 

47 

find  

1 

26 

4 

64 

Lenses  

4 

81 

4 

65 

Letters,  Use  of,  in  algebra  

3 

1 

4 

66 

Light  

4 

72 

4 

57 

'     dispersion  

4 

83 

4 

41 

44     Pencil  of  

4 

73 

55 

44     Propagation  of  

4 

73 

57 

44     Ray  of  

4 

73 

69 

44     Theory  of  

4 

72 

58 

4'      Velocity  of  

4 

73 

61 

Like  numbers  

1 

1 

61 

44     quantities,  Addition  of  

3 

12 

60 

Subtraction    of 

3 

14 

44     terms  

3 

5 

5 

34 

Linear  measure  

2 

8 

5 

48 

44        Surveyors'  

2 

8 

Liquid  bodies  

5 

2 

5 

71 

44       measures  

2 

10 

4 

16 

Liquids,  Specific  heat  of  

4 

61 

Sec. 

Page. 

44        Table  of  specific  grav- 
ity of  

4 

10 

5 

86 

Literal  equations  

3 

56 

4 

4 

44       exponents  

3 

47 

Litmus  

5 

43 

1 

25 

" 

5 

91 

Local  and  relative  values  of  fig- 

1 

47 

ures  

1 

2 

4 

5 

Long-ton  table  

2 

10 

5 

6 

Loss  or  gain  per  cent  

2 

6 

2 

25 

4 

4 

M. 

Sec. 

Page. 

5 

87 

Machines,  Electrostatic  

4 

112 

4 

108 

Magnetic  density  

4 

101 

1 

2 

44        field  

4 

98 

3 

44 

44        force  

4 

98 

2 

47 

44        metals  

4 

98 

2 

50 

44         poles  

4 

97 

2 

43 

Magnetism  

4 

96 

2 

22 

Magnets  

4 

96 

Malleability,  Definition  of  

4 

6 

s*. 

Page. 

Manometer,  The  

4 

27 

3 

52 

Mariotte's  law  

4 

26 

XIV 


INDEX. 


Sec.  Page. 

Mariotte's  law 5  72 

Mass,  Definition  of 4  2 

44  5  2 

44      Motions  of 4  2 

44      of  a  body 4  6 

Matter,  Attraction  of 4  2 

44        Definition  of 4  1 

44 5  1 

Divisions  of 4  3 

44 5  2 

44        Indestructibility  of 5  C 

Motions  of 4  2 

44        Physical  state  of 5  1 

Properties  of 4  4 

Means  (Proportion) 2  47 

Measure,  Cubic 2  9 

Definition  of 2  8 

Dry 2  10 

44          Linear 2  8 

44          of  angles  or  arcs 2  11 

44  money 2  11 

44           44  time 2  11 

44          Square 2  9 

44          Surveyors' linear 2  8 

square 2  9 

Measures  and  weights 5  8 

Classification  of 2  8 

44          Liquid 2  10 

44          Miscellaneous 2  11 

44         of  capacity 2  10 

44           u  extension 2  8 

44           4l  weight 2  9 

44         Standard  units  of 2  8 

Members  of  an  equation 3  51 

Mercurial  barometer 4  22 

Meta-elements 5  30 

Metals  and  non-metals 5  26 

44       Magnetic 4  98 

44       Table  of  conduction  by..  4  56 
44      44  specific   gravity 

of 4  10 

Meter 2  11 

Methyl-orange 5  91 

Metric    system   of  weights  and 

measures 5  8 

Minuend 1  9 

Minus 1  9 

Mixed  number 1  23 

44        To  reduce,  to  im- 
proper fraction  1  25 

44      quantities 3  44 

Mixture,  Mechanical,  and  chemi- 
cal combination 5  5 

Mixtures  of  gases 4  35 

44         Temperature  of 4  62 

Mobility,  Definition  of 4  4 

Molecular  stabil  ity 5  55 

44          volume  . . ,  5  47 


Molecular  weight  and  density. . . 
Molecule,  Definition  of . . . 


Molecules  and  atoms 

44          Chemical  definition  of 

44          Classification  of 

compound,  Classifica- 
tion of 

44          Saturated  and  unsat- 

urated ... 

Money,  Measure  of 

U.  S 

Monomial,  Definition  of 

14  To  extract  root  of 

Monomials,  Addition  of 

Division  of 

Multiplication  of 

Subtraction  of 

Motion,  Atomic 

44        Molecular 

44         of  matter 

Multiplicand 

Multiplication 

in  algebra 

of  decimals 

44  d  enom  in  ate 


Sec. 
5 
4 
5 
5 
5 
5 


Page. 

47 

2 

2 

18 
18 
18 

35 

39 
11 
11 

6 
50 
12 
25 
21 
13 

3 

2 

2 
11 
11 
19 
40 


numbers..  .  .  . 

2 

19 

44   fractions  

1 

31 

u                    n               u 

3 

41 

44   monomials  

3 

21 

44  polynomials... 

3 

21 

Rule  for  

1 

14 

Sign  of  

1 

11 

"              table  

1 

12 

Multiplier  

1 

11 

X. 

Sec. 

Page. 

Nascent  condition  of  elements.. 

5 

87 

Naught  

1 

2 

Negative  and  positive  charges.  .  . 

4 

109 

44          44         quantities 

3 

8 

44         exponent  

3 

48 

Nicholson's  hydrometer  

4 

17 

Nicol's  prism  

4 

92 

Nomenclature  of  acids  and  salts 

5 

43 

44  binary     com- 

pounds   

5 

36 

44  elemental  mole- 

cules     and 

atoms  

5 

19 

Non-conductors,  Electrical  

4 

108 

Non-metals  and  metals  

5 

26 

Normal,  acid,  and  basic  salts  

5 

45 

Notation  

1 

1 

44        Arabic  

1 

2 

44        in  algebra  

3 

3 

Number  

1 

1 

Abstract  

1 

1 

INDEX. 


XV 


Sec. 

Page. 

v 

Sec. 

Page. 

Number,  Concrete  

1 

1 

Pressure,  Absolute  

4 

2? 

"          Denominate  

2 

7 

volume,  and  tempera- 

Mixed   

1 

23 

ture  of  gases  

5 

72 

44         Prime  

1 

20 

Prime  factor  

i 

20 

Unit  of  

1 

1 

"      number  

1 

20 

Numbers,  Like  

1 

1 

Prism,  Foucault's  

4 

93 

Reading  

1 

3 

44       Nicol's  

4 

92 

Unlike  :  

1 

1 

Prisms  

4 

80 

Numeration  

1 

1 

Product  

1 

11 

H 

1 

22 

Propagation  of  heat  

4 

55 

14  light  

4 

72 

0. 

Sec. 

Page. 

Proper  fractions  

1 

23 

Oblique-angled  triangles  

3 

72 

Properties,  Physical  and  chemi- 

Opaque bodies  

4 

72 

cal  

5 

5 

Oxidation  

5 

89 

Proportion  

2 

46 

Oxidizing  agents  

5 

89 

44          Compound  

2 

55 

44           Direct  

2 

47 

P. 

Sec. 

Page. 

how  read  

2 

46 

Parenthesis  

1 

49 

44     written  

2 

46 

Removal  of  

3 

17 

44           Inverse  

2 

47 

Pencil  of  light  

4 

73 

"                  "       

2 

50 

Per  cent.,  Sign  of  

2 

1 

44           Operations  in  

2 

48 

Percentage  

2 

1 

44           Powers  and  roots  in.  . 

2 

52 

"            

2 

2 

44          Rules  for  

2 

47 

composition,    Calcu- 

44          Simple  

2 

55 

lation  of  

5 

64 

Pumps,  Air  

4 

37 

Meaning  of  

2 

1 

Permanent  gas  

4 

3 

Q. 

Sec. 

Page. 

"               " 

5 

2 

Quantities,  Known  and  unknown 

3 

52 

Phenol-phthalein  

5 

91 

Mixed  

3 

44 

Physical  and  chemical  changes  .  . 

5 

4 

44            Positive    and    nega- 

44        properties 

5 

5 

tive  

3 

8 

44        science,  Definition  of... 

4 

1 

Unlike  

8 

14 

Physics,  Province  of  

4 

3 

Quantity,  Definition  of  

3 

3 

Pite,  Voltaic  

4 

120 

41         Factors  of  

3 

4 

Pipettes  

5 

103 

Quotient  

1 

16 

Plane  trigonometry  
Plus  

3 
1 

62 
4 

R. 

Sec. 

Page. 

Polarization  

4 

90 

Radical  sign  

2 

25 

Angle  of  

4 

90 

"          "         

8 

5 

44            Circular  

4 

93 

Radicals,  compoimd,  Names   of 

5 

40 

in  .  opposite    direc- 

Rate (Per  cent)  

2 

2 

tions  

4 

91 

Ratio  

2 

42 

Poles,  Magnetic  

4 

97 

44     Direct  

2 

43 

44      or  electrodes  

4 

118 

44     Inverse  

2 

43 

Polynomials,  Addition  and  sub- 

44    Operations  upon  

2 

45 

traction  of  

3 

15 

44     Reciprocal  

2 

43 

44              Definition  of  

3 

6 

44     Terms  of  

2 

43 

44              Division  of  

3 

26 

44     To  invert  

2 

44 

"              Multiplication  of.. 

3 

21 

44     Value  of  

2 

43 

Porosity,  Definition  of  

4 

4 

Ray  of  light  

4 

73 

Positive  and  negative  charges.  .  . 

4 

109 

Reaction  

5 

90 

''           "           "          quantities 

3 

8 

44         chemical,  Facility  of  — 

5 

57 

Potential  

4 

115 

Reactions  

5 

56 

Power,  Definition  of  

3 

5 

44          Classification  of  

5 

57 

"       of  a  number  

2 

22 

Reagent  

5 

90 

Precipitation  
Prefixes,  Use  of  

5 
5 

58 
37 

Reagents  •  
Reciprocal  of  a  quantity  

5 
3 

56 
34 

XVI 


INDEX. 


Sec, 

Reciprocal  ratio 2 

Reducing  agents 5 

Reduction  of  decimals 1 

"  denominate    num- 
bers   2 

"  "  fractions...  1 


Reflection  and  refraction  ........  4 

"           Laws  of  ...............  4 

Refraction  and  reflection  ........  4 

Double  ...............  4 

Irregular  .............  4 

Law  of  ................  4 

"           Results  of  ............  4 

Table  of  indices  of...  4 

Total  .  .  .  .  !  ............  4 

Relative     and      local    value    of 

figures  ...........................  1 

Remainder  ...........  ,  ..........  .  .  1 

Resistance,  Electrical  ............  4 

Right-angled  triangles  ...........  3 

Root,  Cube  ........................  2 

"      Definition  of  ................  3 

u      Index  of  ....................  2 

"      of  a  number  ................  2 

"      Square  ......................  2 

Roots  of  fractions  ................  2 

"      other    than    square     and 

cube  .......................  2 

Rule  for  addition  .................  1 

"       "    cancelation  .............  1 


compound  proportion..      2 


"    cube  root 
"    division 
''    multiplication 
"    square  root 
"    subtraction 
of  three 


Rules  for  percentage  .............      2 


Page. 
43 
90 
46 

12 
23 
35 
74 
74 


108 
68 
25 
5 
25 
23 
25 
40 

41 
8 
21 
56 
38 
18 
14 
30 
10 
47 
2 


S. 


Sec.  Page. 


Saccharimeter,  The 4  94 

Salt,  Definition  of v 5  42 

Salts,  Acid,  normal,  and  basic. . .  5  45 
"      and    acids,  Nomenclature 

of 5  43 

Saturated      and       unsaturated 

molecules 5  39 

Science,   natural    and    physical, 

Definition  of 4  1 

Score 2  11 

Separation,  Chemical 5  81 

Series,  Electric 4  104 

"       Electromotive 4  119 

Sign  for  dollars 1  48 

"     of  addition 1  4 

'*      "division 1  16 

"      "  equality 1  4 


Sec.  Page. 

Sign  of  multiplication 1  11 

"     "percent 2  1 

"      "subtraction 1  9 

.    "       Radical 2  25 

3  5 

Signs,  Changing,  in  equations...  3  53 

"        used  in  algebra 3  4 

Simple  denominate  number 2  7 

"        equations 3  55 

"       proportion 2  55 

Sine  of  an  angle 3  63 

Sines  and  cosines,  natural,  Table 

of 3  77 

Sodium  carbonate,    Determina- 
tion of 5  95 

Solubility  of  compounds,  Table 

of 5  60 

Solution 5  82 

"         of  right-angled  triangle  3  68 

"          "triangles 3  68 

Solutions,  Standard 5  92 

Specific  gravity,  Definition  of . . .  4  8 
"       of  gases,  Deter- 
mination of...  4  16 
"        of   liquids,   De- 
termination of  4  12 
of  solids,  Deter- 
mination of . . .  4  13 

"              "       Tables  of 4  10 

44        heat 4  58 

44          "    Relation  of,  to 

atomic  weight 5  34 

44           "    Tables  of 4  60 

Spectra,  Dispersion 4  83 

Spectroscope,  The 4  85 

Spectrum,  The 4  83 

Sprengel's  air  pump 4  39 

Square  measure 2  9 

"        Surveyors' 2  9 

"       root 2  25 

"   Proof  of 2  30 

44           "    Rule  for 2  30 

44           "    Short  method  for...  2  31 

Stability,  Molecular 5  55 

Standard  solutions 5  92 

44  units    of    various 

measures 2  8 

Stoichiometry 5  63 

Sublimation 5  86 

Subtraction 1  9 

of  decimals 1  39 

44  denominate  num- 
bers   2  17 

44  fractions 1  29 

3  39 

44  like  quantities.. ..  3  14 

44            "  monomials...  3  13 


INDEX. 


xvii 


Sec.  Page. 

Subtraction  of  polynomials 3  15 

44  unlike  quantities  3  14 

44            Rule  for 1  10 

Sign  of 1  9 

Subtrahend 1  9 

Sulphuric  acid,  Determination  of  5  94 

Supplement  of  an  angle 3  62 

Supporters  of  combustion 5  87 

Surveyors' linear  measure 2  8 

square  measure 2  9 

Symbols  and  formulas 5  24 

44          of  aggregation 1  49 

44             44 3  17 

44             "            Order  of 
prece- 
dence in  1  49 
44  elements,  Table  of..  5  28 
Synthesis  and  analysis 5  7 

T.  Sec.  Page. 

Table  for  conversion  of  English 

measures  into  metric      5       10 
"~     44      conversion  of  metric 
measures  into  Eng- 
lish       5        11 

44        of  atomic  weights 5       28 

44        44  atomicity 5       21 

44        44  coefficients   of   expan- 
sion        4        50 

44        44  conduction  of  heat  by 

metals 4       56 

44        44  electrochemical    char- 
acter of  elements 

44        44  elements 

44        44  indices  of  refraction. .. 

44        44  metric  measures 

44        44  metric  weights 

44        44  natural  sines  and  co- 
sines   

'4        44  natural   tangents   and 

cotangents 3 

44        44  solubility       of       com- 
pounds       5       60 

44        44  temperatures  of  fusion 

and  vaporization 4       67 

44        44  valence 5       23 

Tables  of  specific  gravity  of  sub- 
stances        4        10 

44      trigonometric,  Use  of 3       64 

Tangent  of  an  angle 3        63 

Tangents  and  cotangents,  nat- 
ural, Table  of 3        86 

Temperature 4       42 

Absolute 4       31 

44  Measures  of 4       43 

44  of  mixtures 4        62 

pressure,  and  vol- 
ume of  gases 5        72 


Sec.  Page. 

Temperatures    of     fusion     and 

vaporization,  Table  of  ..........  4  67 

Tenacity,  Definition  of  ...........  4  5 

Tension  of  gases  ..................  4  18 

Terms,  Algebraic  ................  3  5 

"        of  a  fraction  ..............  1  23 

44         '4  an  equation,  Changing 

signs  of  ..............  3  53 

4<        44  a  ratio  .................  2  43 

Ternary  compounds  ..............  5  40 

Theory  of  exponents  .........  ..  ..  3  47 

Thermal  units  ....................  4  58 

Thermometers  ....................  4  43 

Rules    for     com- 

paring ----  ......  4  46 

Time,  Measure  of  .................  2  11 

Torsion  balance  ..................  4  105 

Transformations  in  equations.  .  .  3  52 

Translucent  bodies  ..............  4  72 

Transparent  bodies.  .  '.  ...........  4  72 

Transpositions  in  equations  .....  3  53 

Triangles,  Right-angled  .........  3  68 

44           Solution  of  ............  3  68 

Trigonometric  functions  .........  3  62 

44              tables,  Use  of....  3  64 

Trigonometry,  Plane  ............  3  62 

Trinomial,  Definition  of  .........  3  6 

To  factor  .............  3  30 

Troy  weight  ......................  2  10 


U. 


Sec.  Page. 


5 
5 

31 

28 

44    Hydrogen,  relation    to  air 
unit  ,         

5 

71 

4 

78 

44    of  a  number 

1 

1 

6 

9 

Unknown  quantity  

3 

52 

6 

10 

Unlike  numbers  

1 

1 

44        quantities     

g 

14 

8 

77 

3 

5 

8 

86 

Unsaturated       and       saturated 
molecules..  . 

5 

39 

U.  S.  money 2 

V.  Sec. 

Vacuum,  Definition  of 4 

Valence 5 

44        Graphic  description  of. .  5 

Table  of 5 

Value  of  a  fraction 

44       44  44  ratio 

Vapor 


Vaporization  and   fusion,  Table 
of  temperatures  of 

14  Latent  heat  of 

Vary,  Meaning  of  the  term 

Vinculum 


11 

Page. 
21 
21 


87 


xvm 


INDEX. 


Sec.  Page. 

Voltaic  battery,  The 4  118 

"        elements,  List  of 4  119 

"        or  galvanic  cell 4  118 

pile 4  120 

Volume,  Combination  by 5  50 

Molecular 5  47 

"         temperature,  and  pres- 
sure of  gases 5  72 

AV.                   Sec.  Page. 

Weighing,  Method  of 5  16 

Weight,  Absolute,  of  hydrogen..  5  48 

atomic,  Definition  of....  5  21 
"       Determination 

of ...  5  32 


Sec.  Page. 
Weight,  atomic,  Relation  of,  to 

specific  heat 5  34 

"         Avoirdupois 2  9 

"         Definition  of 4  4 

Laws  of 4  7 

"         Measures  of 2  9 

"         Molecular,  and  density  5  47 

Weights  and  measures,  English  5  12 

"          "             "          Metric..  5  8 

"         atomic,  Table  of 5  28 

Wood,  Table  of  specific  gravity 

of...  4  11 


Zero. 


Sec.  Page. 
1         2 


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